2.1.16 Problem 16

Internal problem ID [10002]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 16
Date solved : Friday, January 30, 2026 at 04:33:15 PM
CAS classification : [[_homogeneous, `class G`], _rational, [_Abel, `2nd type`, `class B`]]

2.1.16.1 Solved using first_order_ode_exact

1.194 (sec)

Entering first order ode exact solver

\begin{align*} y^{\prime }&=\frac {-y x -1}{4 x^{3} y-2 x^{2}} \\ \end{align*}

To solve an ode of the form

\begin{equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A}\end{equation}

We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives

\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \]

Hence

\begin{equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B}\end{equation}

Comparing (A,B) shows that

\begin{align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end{align*}

But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that

\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\]

If the above condition is satisfied, then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not satisfied then this method will not work and we have to now look for an integrating factor to force this condition, which might or might not exist. The first step is to write the ODE in standard form to check for exactness, which is

\[ M(x,y) \mathop {\mathrm {d}x}+ N(x,y) \mathop {\mathrm {d}y}=0 \tag {1A} \]

Therefore

\begin{align*} \mathop {\mathrm {d}y} &= \left (\frac {-x y -1}{4 x^{3} y -2 x^{2}}\right )\mathop {\mathrm {d}x}\\ \left (-\frac {-x y -1}{4 x^{3} y -2 x^{2}}\right ) \mathop {\mathrm {d}x} + \mathop {\mathrm {d}y} &= 0 \tag {2A} \end{align*}

Comparing (1A) and (2A) shows that

\begin{align*} M(x,y) &= -\frac {-x y -1}{4 x^{3} y -2 x^{2}}\\ N(x,y) &= 1 \end{align*}

The next step is to determine if the ODE is is exact or not. The ODE is exact when the following condition is satisfied

\[ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x} \]

Using result found above gives

\begin{align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (-\frac {-x y -1}{4 x^{3} y -2 x^{2}}\right )\\ &= -\frac {3}{2 x \left (2 x y -1\right )^{2}} \end{align*}

And

\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (1\right )\\ &= 0 \end{align*}

Since \(\frac {\partial M}{\partial y} \neq \frac {\partial N}{\partial x}\), then the ODE is not exact. Since the ODE is not exact, we will try to find an integrating factor to make it exact. Let

\begin{align*} A &= \frac {1}{N} \left (\frac {\partial M}{\partial y} - \frac {\partial N}{\partial x} \right ) \\ &=1\left ( \left ( \frac {x}{4 x^{3} y -2 x^{2}}+\frac {4 \left (-x y -1\right ) x^{3}}{\left (4 x^{3} y -2 x^{2}\right )^{2}}\right ) - \left (0 \right ) \right ) \\ &=-\frac {3}{2 x \left (2 x y -1\right )^{2}} \end{align*}

Since \(A\) depends on \(y\), it can not be used to obtain an integrating factor. We will now try a second method to find an integrating factor. Let

\begin{align*} B &= \frac {1}{M} \left ( \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} \right ) \\ &=\frac {4 x^{3} y -2 x^{2}}{x y +1}\left ( \left ( 0\right ) - \left (\frac {x}{4 x^{3} y -2 x^{2}}+\frac {4 \left (-x y -1\right ) x^{3}}{\left (4 x^{3} y -2 x^{2}\right )^{2}} \right ) \right ) \\ &=\frac {3 x}{2 x^{2} y^{2}+x y -1} \end{align*}

Since \(B\) depends on \(x\), it can not be used to obtain an integrating factor.We will now try a third method to find an integrating factor. Let

\[ R = \frac { \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} } {x M - y N} \]

\(R\) is now checked to see if it is a function of only \(t=x y\). Therefore

\begin{align*} R &= \frac { \frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} } {x M - y N} \\ &= \frac {\left (0\right )-\left (\frac {x}{4 x^{3} y -2 x^{2}}+\frac {4 \left (-x y -1\right ) x^{3}}{\left (4 x^{3} y -2 x^{2}\right )^{2}}\right )} {x\left (-\frac {-x y -1}{4 x^{3} y -2 x^{2}}\right ) - y\left (1\right )} \\ &= -\frac {3}{8 x^{3} y^{3}-10 x^{2} y^{2}+x y +1} \end{align*}

Replacing all powers of terms \(x y\) by \(t\) gives

\[ R = -\frac {3}{8 t^{3}-10 t^{2}+t +1} \]

Since \(R\) depends on \(t\) only, then it can be used to find an integrating factor. Let the integrating factor be \(\mu \) then

\begin{align*} \mu &= e^{\int R \mathop {\mathrm {d}t}} \\ &= e^{\int \left (-\frac {3}{8 t^{3}-10 t^{2}+t +1}\right )\mathop {\mathrm {d}t} } \end{align*}

The result of integrating gives

\begin{align*} \mu &= e^{-\frac {2 \ln \left (4 t +1\right )}{5}+\ln \left (2 t -1\right )-\frac {3 \ln \left (t -1\right )}{5} } \\ &= \frac {2 t -1}{\left (4 t +1\right )^{{2}/{5}} \left (t -1\right )^{{3}/{5}}} \end{align*}

Now \(t\) is replaced back with \(x y\) giving

\[ \mu =\frac {2 x y -1}{\left (4 x y +1\right )^{{2}/{5}} \left (x y -1\right )^{{3}/{5}}} \]

Multiplying \(M\) and \(N\) by this integrating factor gives new \(M\) and new \(N\) which are called \( \overline {M}\) and \( \overline {N}\) so not to confuse them with the original \(M\) and \(N\)

\begin{align*} \overline {M} &=\mu M \\ &= \frac {2 x y -1}{\left (4 x y +1\right )^{{2}/{5}} \left (x y -1\right )^{{3}/{5}}}\left (-\frac {-x y -1}{4 x^{3} y -2 x^{2}}\right ) \\ &= \frac {x y +1}{2 \left (x y -1\right )^{{3}/{5}} \left (4 x y +1\right )^{{2}/{5}} x^{2}} \end{align*}

And

\begin{align*} \overline {N} &=\mu N \\ &= \frac {2 x y -1}{\left (4 x y +1\right )^{{2}/{5}} \left (x y -1\right )^{{3}/{5}}}\left (1\right ) \\ &= \frac {2 x y -1}{\left (4 x y +1\right )^{{2}/{5}} \left (x y -1\right )^{{3}/{5}}} \end{align*}

A modified ODE is now obtained from the original ODE, which is exact and can solved. The modified ODE is

\begin{align*} \overline {M} + \overline {N} \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \\ \left (\frac {x y +1}{2 \left (x y -1\right )^{{3}/{5}} \left (4 x y +1\right )^{{2}/{5}} x^{2}}\right ) + \left (\frac {2 x y -1}{\left (4 x y +1\right )^{{2}/{5}} \left (x y -1\right )^{{3}/{5}}}\right ) \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}} &= 0 \end{align*}

The following equations are now set up to solve for the function \(\phi \left (x,y\right )\)

\begin{align*} \frac {\partial \phi }{\partial x } &= \overline {M}\tag {1} \\ \frac {\partial \phi }{\partial y } &= \overline {N}\tag {2} \end{align*}

Integrating (1) w.r.t. \(x\) gives

\begin{align*} \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int \overline {M}\mathop {\mathrm {d}x} \\ \int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int \frac {x y +1}{2 \left (x y -1\right )^{{3}/{5}} \left (4 x y +1\right )^{{2}/{5}} x^{2}}\mathop {\mathrm {d}x} \\ \tag{3} \phi &= \frac {\left (x y -1\right )^{{2}/{5}} \left (4 x y +1\right )^{{3}/{5}}}{2 x}+ f(y) \\ \end{align*}

Where \(f(y)\) is used for the constant of integration since \(\phi \) is a function of both \(x\) and \(y\). Taking derivative of equation (3) w.r.t \(y\) gives

\begin{align*} \tag{4} \frac {\partial \phi }{\partial y} &= \frac {\left (4 x y +1\right )^{{3}/{5}}}{5 \left (x y -1\right )^{{3}/{5}}}+\frac {6 \left (x y -1\right )^{{2}/{5}}}{5 \left (4 x y +1\right )^{{2}/{5}}}+f'(y) \\ &=\frac {2 x y -1}{\left (4 x y +1\right )^{{2}/{5}} \left (x y -1\right )^{{3}/{5}}}+f'(y) \\ \end{align*}

But equation (2) says that \(\frac {\partial \phi }{\partial y} = \frac {2 x y -1}{\left (4 x y +1\right )^{{2}/{5}} \left (x y -1\right )^{{3}/{5}}}\). Therefore equation (4) becomes

\begin{equation} \tag{5} \frac {2 x y -1}{\left (4 x y +1\right )^{{2}/{5}} \left (x y -1\right )^{{3}/{5}}} = \frac {2 x y -1}{\left (4 x y +1\right )^{{2}/{5}} \left (x y -1\right )^{{3}/{5}}}+f'(y) \end{equation}

Solving equation (5) for \( f'(y)\) gives

\[ f'(y) = 0 \]

Therefore

\[ f(y) = c_1 \]

Where \(c_1\) is constant of integration. Substituting this result for \(f(y)\) into equation (3) gives \(\phi \)

\[ \phi = \frac {\left (x y -1\right )^{{2}/{5}} \left (4 x y +1\right )^{{3}/{5}}}{2 x}+ c_1 \]

But since \(\phi \) itself is a constant function, then let \(\phi =c_2\) where \(c_2\) is new constant and combining \(c_1\) and \(c_2\) constants into the constant \(c_1\) gives the solution as

\[ c_1 = \frac {\left (x y -1\right )^{{2}/{5}} \left (4 x y +1\right )^{{3}/{5}}}{2 x} \]

Direction field \(y^{\prime } = \frac {-y x -1}{4 x^{3} y-2 x^{2}}\)	Isoclines for \(y^{\prime } = \frac {-y x -1}{4 x^{3} y-2 x^{2}}\)

Summary of solutions found

\begin{align*} \frac {\left (4 y x +1\right )^{{3}/{5}} \left (y x -1\right )^{{2}/{5}}}{2 x} &= c_1 \\ \end{align*}

2.1.16.2 Solved using first_order_ode_isobaric

1.918 (sec)

Entering first order ode isobaric solver

\begin{align*} y^{\prime }&=\frac {-y x -1}{4 x^{3} y-2 x^{2}} \\ \end{align*}

Solving for \(y'\) gives

\[ y' = \frac {-y x -1}{4 x^{3} y-2 x^{2}} \]

An ode \(y^{\prime }=f(x,y)\) is isobaric if

\[ f(t x, t^m y) = t^{m-1} f(x,y)\tag {1} \]

Where here

\[ f(x,y) = \frac {-y x -1}{4 x^{3} y-2 x^{2}}\tag {2} \]

\(m\) is the order of isobaric. Substituting (2) into (1) and solving for \(m\) gives

\[ m = -1 \]

Since the ode is isobaric of order \(m=-1\), then the substitution

\begin{align*} y&=u x^m \\ &=\frac {u}{x} \end{align*}

Converts the ODE to a separable in \(u \left (x \right )\). Performing this substitution gives

\[ -\frac {u \left (x \right )}{x^{2}}+\frac {u^{\prime }\left (x \right )}{x} = \frac {-u \left (x \right )-1}{4 x^{2} u \left (x \right )-2 x^{2}} \]

Entering first order ode separable solverThe ode

\begin{equation} u^{\prime }\left (x \right ) = \frac {\left (4 u \left (x \right )+1\right ) \left (u \left (x \right )-1\right )}{2 x \left (2 u \left (x \right )-1\right )} \end{equation}

is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= \frac {\left (4 u \left (x \right )+1\right ) \left (u \left (x \right )-1\right )}{2 x \left (2 u \left (x \right )-1\right )}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{2 x}\\ g(u) &= \frac {\left (u -1\right ) \left (4 u +1\right )}{2 u -1} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\ \int { \frac {2 u -1}{\left (u -1\right ) \left (4 u +1\right )}\,du} &= \int { \frac {1}{2 x} \,dx} \\ \end{align*}

\[ \frac {3 \ln \left (4 u \left (x \right )+1\right )}{10}+\frac {\ln \left (u \left (x \right )-1\right )}{5}=\ln \left (\sqrt {x}\right )+c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ \frac {\left (u -1\right ) \left (4 u +1\right )}{2 u -1}=0 \]

for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=1\\ u \left (x \right )&=-{\frac {1}{4}} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \frac {3 \ln \left (4 u \left (x \right )+1\right )}{10}+\frac {\ln \left (u \left (x \right )-1\right )}{5} &= \ln \left (\sqrt {x}\right )+c_1 \\ u \left (x \right ) &= 1 \\ u \left (x \right ) &= -{\frac {1}{4}} \\ \end{align*}

Converting \(\frac {3 \ln \left (4 u \left (x \right )+1\right )}{10}+\frac {\ln \left (u \left (x \right )-1\right )}{5} = \frac {\ln \left (x \right )}{2}+c_1\) back to \(y\) gives

\begin{align*} \frac {\ln \left (y x -1\right )}{5}+\frac {3 \ln \left (4 y x +1\right )}{10} = \frac {\ln \left (x \right )}{2}+c_1 \end{align*}

Converting \(u \left (x \right ) = 1\) back to \(y\) gives

\begin{align*} y x = 1 \end{align*}

Converting \(u \left (x \right ) = -{\frac {1}{4}}\) back to \(y\) gives

\begin{align*} y x = -{\frac {1}{4}} \end{align*}

Solving for \(y\) gives

\begin{align*} \frac {\ln \left (y x -1\right )}{5}+\frac {3 \ln \left (4 y x +1\right )}{10} &= \frac {\ln \left (x \right )}{2}+c_1 \\ y &= \frac {1}{x} \\ y &= -\frac {1}{4 x} \\ \end{align*}

Direction field \(y^{\prime } = \frac {-y x -1}{4 x^{3} y-2 x^{2}}\)	Isoclines for \(y^{\prime } = \frac {-y x -1}{4 x^{3} y-2 x^{2}}\)

Summary of solutions found

\begin{align*} \frac {\ln \left (y x -1\right )}{5}+\frac {3 \ln \left (4 y x +1\right )}{10} &= \frac {\ln \left (x \right )}{2}+c_1 \\ y &= \frac {1}{x} \\ y &= -\frac {1}{4 x} \\ \end{align*}

2.1.16.3 Solved using first_order_ode_homog_type_G

1.760 (sec)

Entering first order ode homog type G solver

\begin{align*} y^{\prime }&=\frac {-y x -1}{4 x^{3} y-2 x^{2}} \\ \end{align*}

Multiplying the right side of the ode, which is \(\frac {-x y -1}{4 x^{3} y -2 x^{2}}\) by \(\frac {x}{y}\) gives

\begin{align*} y^{\prime } &= \left (\frac {x}{y}\right ) \frac {-x y -1}{4 x^{3} y -2 x^{2}}\\ &= -\frac {x y +1}{2 x y \left (2 x y -1\right )}\\ &= F(x,y) \end{align*}

Since \(F \left (x , y\right )\) has \(y\), then let

\begin{align*} f_x&= x \left (\frac {\partial }{\partial x}F \left (x , y\right )\right )\\ &= \frac {2 x^{2} y^{2}+4 x y -1}{2 x y \left (2 x y -1\right )^{2}}\\ f_y&= y \left (\frac {\partial }{\partial y}F \left (x , y\right )\right )\\ &= \frac {2 x^{2} y^{2}+4 x y -1}{2 x y \left (2 x y -1\right )^{2}}\\ \alpha &= \frac {f_x}{f_y} \\ &=1 \end{align*}

Since \(\alpha \) is independent of \(x,y\) then this is Homogeneous type G.

Let

\begin{align*} y&=\frac {z}{x^ \alpha }\\ &=\frac {z}{x} \end{align*}

Substituting the above back into \(F(x,y)\) gives

\begin{align*} F \left (z \right ) &=-\frac {z +1}{2 z \left (2 z -1\right )} \end{align*}

We see that \(F \left (z \right )\) does not depend on \(x\) nor on \(y\). If this was not the case, then this method will not work.

Therefore, the implicit solution is given by

\begin{align*} \ln \left (x \right )- c_1 - \int ^{y x^\alpha } \frac {1}{z \left (\alpha + F(z)\right ) } \,dz & = 0 \end{align*}

Which gives

\[ \ln \left (x \right )-c_1 +\int _{}^{y x}\frac {1}{z \left (-1+\frac {z +1}{2 z \left (2 z -1\right )}\right )}d z = 0 \]

The value of the above is

\[ \ln \left (x \right )-c_1 -\frac {2 \ln \left (y x -1\right )}{5}-\frac {3 \ln \left (4 y x +1\right )}{5} = 0 \]

Direction field \(y^{\prime } = \frac {-y x -1}{4 x^{3} y-2 x^{2}}\)	Isoclines for \(y^{\prime } = \frac {-y x -1}{4 x^{3} y-2 x^{2}}\)

Summary of solutions found

\begin{align*} \ln \left (x \right )-c_1 -\frac {2 \ln \left (y x -1\right )}{5}-\frac {3 \ln \left (4 y x +1\right )}{5} &= 0 \\ \end{align*}

Entering first order ode abel second kind solver

\begin{align*} y^{\prime }&=\frac {-y x -1}{4 x^{3} y-2 x^{2}} \\ \end{align*}

2.1.16.4 Solved using first_order_ode_abel_second_kind_case_3

2.336 (sec)

Applying transformation

\begin{align*} y&=\frac {1}{u(x)}-g \end{align*}

Results in the new ode which is Abel first kind

\begin{align*} -\frac {u \left (x \right )}{x^{2}}+\frac {u^{\prime }\left (x \right )}{x} = \frac {-u \left (x \right )-1}{4 x^{2} u \left (x \right )-2 x^{2}} \end{align*}

Which is now solved. Entering first order ode separable solverThe ode

\begin{equation} u^{\prime }\left (x \right ) = \frac {\left (4 u \left (x \right )+1\right ) \left (u \left (x \right )-1\right )}{2 x \left (2 u \left (x \right )-1\right )} \end{equation}

is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= \frac {\left (4 u \left (x \right )+1\right ) \left (u \left (x \right )-1\right )}{2 x \left (2 u \left (x \right )-1\right )}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{2 x}\\ g(u) &= \frac {\left (4 u +1\right ) \left (u -1\right )}{2 u -1} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\ \int { \frac {2 u -1}{\left (4 u +1\right ) \left (u -1\right )}\,du} &= \int { \frac {1}{2 x} \,dx} \\ \end{align*}

\[ \frac {3 \ln \left (4 u \left (x \right )+1\right )}{10}+\frac {\ln \left (u \left (x \right )-1\right )}{5}=\ln \left (\sqrt {x}\right )+c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ \frac {\left (4 u +1\right ) \left (u -1\right )}{2 u -1}=0 \]

for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=1\\ u \left (x \right )&=-{\frac {1}{4}} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \frac {3 \ln \left (4 u \left (x \right )+1\right )}{10}+\frac {\ln \left (u \left (x \right )-1\right )}{5} &= \ln \left (\sqrt {x}\right )+c_1 \\ u \left (x \right ) &= 1 \\ u \left (x \right ) &= -{\frac {1}{4}} \\ \end{align*}

Substituting \(u \left (x \right )=\frac {1}{y-\frac {1}{2 x}}\) in the above solution gives

\begin{align*} \frac {\ln \left (y x -1\right )}{5}+\frac {3 \ln \left (4 y x +1\right )}{10} = \frac {\ln \left (x \right )}{2}+c_1 \end{align*}

Now we transform the solution \(u \left (x \right ) = 1\) to \(y\) using \(u \left (x \right )=\frac {1}{y-\frac {1}{2 x}}\) which gives

\begin{align*} y &= \frac {1}{u \left (x \right )} - -\frac {1}{2 x}\\ &= 1+\frac {1}{2 x} \end{align*}

Now we transform the solution \(u \left (x \right ) = -{\frac {1}{4}}\) to \(y\) using \(u \left (x \right )=\frac {1}{y-\frac {1}{2 x}}\) which gives

\begin{align*} y &= \frac {1}{u \left (x \right )} - -\frac {1}{2 x}\\ &= -4+\frac {1}{2 x} \end{align*}

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

Direction field \(y^{\prime } = \frac {-y x -1}{4 x^{3} y-2 x^{2}}\)	Isoclines for \(y^{\prime } = \frac {-y x -1}{4 x^{3} y-2 x^{2}}\)

Summary of solutions found

\begin{align*} \frac {\ln \left (y x -1\right )}{5}+\frac {3 \ln \left (4 y x +1\right )}{10} &= \frac {\ln \left (x \right )}{2}+c_1 \\ \end{align*}

2.1.16.5 Solved using first_order_ode_abel_second_kind_solved_by_converting_to_first_kind

215.674 (sec)

This is Abel second kind ODE, it has the form

\[ \left (y+g\right )y^{\prime }= f_0(x)+f_1(x) y +f_2(x)y^{2}+f_3(x)y^{3} \]

Comparing the above to given ODE which is

\begin{align*}y^{\prime } = \frac {-y x -1}{4 x^{3} y-2 x^{2}}\tag {1} \end{align*}

Shows that

\begin{align*} g &= -\frac {1}{2 x}\\ f_0 &= -\frac {1}{4 x^{3}}\\ f_1 &= -\frac {1}{4 x^{2}}\\ f_2 &= 0\\ f_3 &= 0 \end{align*}

Applying transformation

\begin{align*} y&=\frac {1}{u(x)}-g \end{align*}

Results in the new ode which is Abel first kind

\begin{align*} u^{\prime }\left (x \right ) = \frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )^{2}}{4 x^{2}} \end{align*}

Which is now solved.

2.1.16.6 Solved using first_order_ode_isobaric

1.946 (sec)

Entering first order ode isobaric solverSolving for \(u'\) gives

\[ u' = \frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )^{2}}{4 x^{2}} \]

An ode \(u^{\prime }\left (x \right )=f(x,u)\) is isobaric if

\[ f(t x, t^m u) = t^{m-1} f(x,u)\tag {1} \]

Where here

\[ f(x,u) = \frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )^{2}}{4 x^{2}}\tag {2} \]

\(m\) is the order of isobaric. Substituting (2) into (1) and solving for \(m\) gives

\[ m = 1 \]

Since the ode is isobaric of order \(m=1\), then the substitution

\begin{align*} u&=u x^m \\ &=u x \end{align*}

Converts the ODE to a separable in \(u \left (x \right )\). Performing this substitution gives

\[ u \left (x \right )+x u^{\prime }\left (x \right ) = \frac {3 u \left (x \right )^{3}}{8}-\frac {u \left (x \right )^{2}}{4} \]

Entering first order ode separable solverThe ode

\begin{equation} u^{\prime }\left (x \right ) = \frac {u \left (x \right ) \left (3 u \left (x \right )+4\right ) \left (u \left (x \right )-2\right )}{8 x} \end{equation}

is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= \frac {u \left (x \right ) \left (3 u \left (x \right )+4\right ) \left (u \left (x \right )-2\right )}{8 x}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{8 x}\\ g(u) &= u \left (3 u +4\right ) \left (u -2\right ) \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\ \int { \frac {1}{u \left (3 u +4\right ) \left (u -2\right )}\,du} &= \int { \frac {1}{8 x} \,dx} \\ \end{align*}

\[ -\frac {\ln \left (u \left (x \right )\right )}{8}+\frac {\ln \left (u \left (x \right )-2\right )}{20}+\frac {3 \ln \left (3 u \left (x \right )+4\right )}{40}=\ln \left (x^{{1}/{8}}\right )+c_2 \]

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ u \left (3 u +4\right ) \left (u -2\right )=0 \]

for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=0\\ u \left (x \right )&=2 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} -\frac {\ln \left (u \left (x \right )\right )}{8}+\frac {\ln \left (u \left (x \right )-2\right )}{20}+\frac {3 \ln \left (3 u \left (x \right )+4\right )}{40} &= \ln \left (x^{{1}/{8}}\right )+c_2 \\ u \left (x \right ) &= 0 \\ u \left (x \right ) &= 2 \\ \end{align*}

Converting \(-\frac {\ln \left (u \left (x \right )\right )}{8}+\frac {\ln \left (u \left (x \right )-2\right )}{20}+\frac {3 \ln \left (3 u \left (x \right )+4\right )}{40} = \frac {\ln \left (x \right )}{8}+c_2\) back to \(u \left (x \right )\) gives

\begin{align*} -\frac {\ln \left (\frac {u \left (x \right )}{x}\right )}{8}+\frac {\ln \left (\frac {u \left (x \right )}{x}-2\right )}{20}+\frac {3 \ln \left (\frac {3 u \left (x \right )}{x}+4\right )}{40} = \frac {\ln \left (x \right )}{8}+c_2 \end{align*}

Converting \(u \left (x \right ) = 0\) back to \(u \left (x \right )\) gives

\begin{align*} \frac {u \left (x \right )}{x} = 0 \end{align*}

Converting \(u \left (x \right ) = 2\) back to \(u \left (x \right )\) gives

\begin{align*} \frac {u \left (x \right )}{x} = 2 \end{align*}

Simplifying the above gives

\begin{align*} -\frac {\ln \left (\frac {u \left (x \right )}{x}\right )}{8}+\frac {\ln \left (\frac {u \left (x \right )-2 x}{x}\right )}{20}+\frac {3 \ln \left (\frac {3 u \left (x \right )+4 x}{x}\right )}{40} &= \frac {\ln \left (x \right )}{8}+c_2 \\ \frac {u \left (x \right )}{x} &= 0 \\ \frac {u \left (x \right )}{x} &= 2 \\ \end{align*}

Solving for \(u \left (x \right )\) gives

\begin{align*} -\frac {\ln \left (\frac {u \left (x \right )}{x}\right )}{8}+\frac {\ln \left (\frac {u \left (x \right )-2 x}{x}\right )}{20}+\frac {3 \ln \left (\frac {3 u \left (x \right )+4 x}{x}\right )}{40} &= \frac {\ln \left (x \right )}{8}+c_2 \\ u \left (x \right ) &= 0 \\ u \left (x \right ) &= 2 x \\ \end{align*}

Summary of solutions found

\begin{align*} -\frac {\ln \left (\frac {u \left (x \right )}{x}\right )}{8}+\frac {\ln \left (\frac {u \left (x \right )-2 x}{x}\right )}{20}+\frac {3 \ln \left (\frac {3 u \left (x \right )+4 x}{x}\right )}{40} &= \frac {\ln \left (x \right )}{8}+c_2 \\ u \left (x \right ) &= 0 \\ u \left (x \right ) &= 2 x \\ \end{align*}

2.1.16.7 Solved using first_order_ode_homog_A

2.985 (sec)

Entering first order ode homog A solver

\begin{align*} u^{\prime }\left (x \right )&=\frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )^{2}}{4 x^{2}} \\ \end{align*}

In canonical form, the ODE is

\begin{align*} u' &= F(x,u)\\ &= \frac {3 u^{3}}{8 x^{3}}-\frac {u^{2}}{4 x^{2}}\tag {1} \end{align*}

An ode of the form \(u' = \frac {M(x,u)}{N(x,u)}\) is called homogeneous if the functions \(M(x,u)\) and \(N(x,u)\) are both homogeneous functions and of the same order. Recall that a function \(f(x,u)\) is homogeneous of order \(n\) if

\[ f(t^n x, t^n u)= t^n f(x,u) \]

In this case, it can be seen that both \(M=u^{2} \left (3 u -2 x \right )\) and \(N=8 x^{3}\) are both homogeneous and of the same order \(n=3\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {u}{x}\), or \(u=ux\). Hence

\[ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u \]

Applying the transformation \(u=ux\) to the above ODE in (1) gives

\begin{align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u &= \frac {3}{8} u^{3}-\frac {1}{4} u^{2}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {\frac {3 u \left (x \right )^{3}}{8}-\frac {u \left (x \right )^{2}}{4}-u \left (x \right )}{x} \end{align*}

Or

\[ u^{\prime }\left (x \right )-\frac {\frac {3 u \left (x \right )^{3}}{8}-\frac {u \left (x \right )^{2}}{4}-u \left (x \right )}{x} = 0 \]

Or

\[ -3 u \left (x \right )^{3}+8 u^{\prime }\left (x \right ) x +2 u \left (x \right )^{2}+8 u \left (x \right ) = 0 \]

Which is now solved as separable in \(u \left (x \right )\).

The ode

\begin{equation} u^{\prime }\left (x \right ) = \frac {u \left (x \right ) \left (3 u \left (x \right )+4\right ) \left (u \left (x \right )-2\right )}{8 x} \end{equation}

is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= \frac {u \left (x \right ) \left (3 u \left (x \right )+4\right ) \left (u \left (x \right )-2\right )}{8 x}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{8 x}\\ g(u) &= u \left (3 u +4\right ) \left (u -2\right ) \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\ \int { \frac {1}{u \left (3 u +4\right ) \left (u -2\right )}\,du} &= \int { \frac {1}{8 x} \,dx} \\ \end{align*}

\[ -\frac {\ln \left (u \left (x \right )\right )}{8}+\frac {\ln \left (u \left (x \right )-2\right )}{20}+\frac {3 \ln \left (3 u \left (x \right )+4\right )}{40}=\ln \left (x^{{1}/{8}}\right )+c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ u \left (3 u +4\right ) \left (u -2\right )=0 \]

for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=0\\ u \left (x \right )&=2 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} -\frac {\ln \left (u \left (x \right )\right )}{8}+\frac {\ln \left (u \left (x \right )-2\right )}{20}+\frac {3 \ln \left (3 u \left (x \right )+4\right )}{40} &= \ln \left (x^{{1}/{8}}\right )+c_1 \\ u \left (x \right ) &= 0 \\ u \left (x \right ) &= 2 \\ \end{align*}

Converting \(-\frac {\ln \left (u \left (x \right )\right )}{8}+\frac {\ln \left (u \left (x \right )-2\right )}{20}+\frac {3 \ln \left (3 u \left (x \right )+4\right )}{40} = \ln \left (x^{{1}/{8}}\right )+c_1\) back to \(u \left (x \right )\) gives

\begin{align*} -\ln \left (\frac {u \left (x \right )}{x}\right )+\frac {2 \ln \left (\frac {u \left (x \right )-2 x}{x}\right )}{5}+\frac {3 \ln \left (\frac {3 u \left (x \right )+4 x}{x}\right )}{5} = \ln \left (x \right )+8 c_1 \end{align*}

Converting \(u \left (x \right ) = 0\) back to \(u \left (x \right )\) gives

\begin{align*} u \left (x \right ) = 0 \end{align*}

Converting \(u \left (x \right ) = 2\) back to \(u \left (x \right )\) gives

\begin{align*} u \left (x \right ) = 2 x \end{align*}

Summary of solutions found

\begin{align*} -\ln \left (\frac {u \left (x \right )}{x}\right )+\frac {2 \ln \left (\frac {u \left (x \right )-2 x}{x}\right )}{5}+\frac {3 \ln \left (\frac {3 u \left (x \right )+4 x}{x}\right )}{5} &= \ln \left (x \right )+8 c_1 \\ u \left (x \right ) &= 0 \\ u \left (x \right ) &= 2 x \\ \end{align*}

2.1.16.8 Solved using first_order_ode_dAlembert

70.368 (sec)

Entering first order ode dAlembert solver

\begin{align*} u^{\prime }\left (x \right )&=\frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )^{2}}{4 x^{2}} \\ \end{align*}

Let \(p=u^{\prime }\left (x \right )\) the ode becomes

\begin{align*} p = \frac {3 u^{3}}{8 x^{3}}-\frac {u^{2}}{4 x^{2}} \end{align*}

Solving for \(u \left (x \right )\) from the above results in

\begin{align*} \tag{1} u \left (x \right ) &= \left (\frac {\left (8+972 p +36 \sqrt {729 p^{2}+12 p}\right )^{{1}/{3}}}{9}+\frac {4}{9 \left (8+972 p +36 \sqrt {729 p^{2}+12 p}\right )^{{1}/{3}}}+\frac {2}{9}\right ) x \\ \tag{2} u \left (x \right ) &= \left (-\frac {\left (8+972 p +36 \sqrt {729 p^{2}+12 p}\right )^{{1}/{3}}}{18}-\frac {2}{9 \left (8+972 p +36 \sqrt {729 p^{2}+12 p}\right )^{{1}/{3}}}+\frac {2}{9}+i \sqrt {3}\, \left (\frac {\left (8+972 p +36 \sqrt {729 p^{2}+12 p}\right )^{{1}/{3}}}{18}-\frac {2}{9 \left (8+972 p +36 \sqrt {729 p^{2}+12 p}\right )^{{1}/{3}}}\right )\right ) x \\ \tag{3} u \left (x \right ) &= \left (-\frac {\left (8+972 p +36 \sqrt {729 p^{2}+12 p}\right )^{{1}/{3}}}{18}-\frac {2}{9 \left (8+972 p +36 \sqrt {729 p^{2}+12 p}\right )^{{1}/{3}}}+\frac {2}{9}-i \sqrt {3}\, \left (\frac {\left (8+972 p +36 \sqrt {729 p^{2}+12 p}\right )^{{1}/{3}}}{18}-\frac {2}{9 \left (8+972 p +36 \sqrt {729 p^{2}+12 p}\right )^{{1}/{3}}}\right )\right ) x \\ \end{align*}

This has the form

\begin{align*} u=x f(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=u'(x)\). Each of the above ode’s is dAlembert ode which is now solved.

Solving ode 1A

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(u \left (x \right )=x f + g\) to (1A) shows that

\begin{align*} f &= \frac {\left (8+972 p +36 \sqrt {3}\, \sqrt {p \left (4+243 p \right )}\right )^{{1}/{3}}}{9}+\frac {4}{9 \left (8+972 p +36 \sqrt {3}\, \sqrt {p \left (4+243 p \right )}\right )^{{1}/{3}}}+\frac {2}{9}\\ g &= 0 \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p -\frac {\left (8+972 p +36 \sqrt {3}\, \sqrt {p \left (4+243 p \right )}\right )^{{1}/{3}}}{9}-\frac {4}{9 \left (8+972 p +36 \sqrt {3}\, \sqrt {p \left (4+243 p \right )}\right )^{{1}/{3}}}-\frac {2}{9} = \left (\frac {36 x}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}}}+\frac {8 x \sqrt {3}}{3 \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}} \sqrt {243 p^{2}+4 p}}+\frac {324 x \sqrt {3}\, p}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}} \sqrt {243 p^{2}+4 p}}-\frac {144 x}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{4}/{3}}}-\frac {32 x \sqrt {3}}{3 \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{4}/{3}} \sqrt {243 p^{2}+4 p}}-\frac {1296 x \sqrt {3}\, p}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{4}/{3}} \sqrt {243 p^{2}+4 p}}\right ) p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p -\frac {\left (8+972 p +36 \sqrt {3}\, \sqrt {p \left (4+243 p \right )}\right )^{{1}/{3}}}{9}-\frac {4}{9 \left (8+972 p +36 \sqrt {3}\, \sqrt {p \left (4+243 p \right )}\right )^{{1}/{3}}}-\frac {2}{9} = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=2 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} u \left (x \right ) = 2 x \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )-\frac {\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )^{{1}/{3}}}{9}-\frac {4}{9 \left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )^{{1}/{3}}}-\frac {2}{9}}{\frac {36 x}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{2}/{3}}}+\frac {8 x \sqrt {3}}{3 \left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{2}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}+\frac {324 x \sqrt {3}\, p \left (x \right )}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{2}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}-\frac {144 x}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{4}/{3}}}-\frac {32 x \sqrt {3}}{3 \left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{4}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}-\frac {1296 x \sqrt {3}\, p \left (x \right )}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{4}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

The ode

\begin{equation} p^{\prime }\left (x \right ) = \frac {\sqrt {3}\, \left (-9 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} p \left (x \right )+2 \,6^{{2}/{3}}+2 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}}{18 \left (6^{{1}/{3}}-\left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}}\right ) \left (6^{{1}/{3}}+\left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}}\right ) x} \end{equation}

is separable as it can be written as

\begin{align*} p^{\prime }\left (x \right )&= \frac {\sqrt {3}\, \left (-9 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} p \left (x \right )+2 \,6^{{2}/{3}}+2 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}}{18 \left (6^{{1}/{3}}-\left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}}\right ) \left (6^{{1}/{3}}+\left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}}\right ) x}\\ &= f(x) g(p) \end{align*}

Where

\begin{align*} f(x) &= \frac {\sqrt {3}}{18 x}\\ g(p) &= \frac {\left (-9 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} p +2 \,6^{{2}/{3}}+2 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{2}/{3}}\right ) \sqrt {p \left (4+243 p \right )}}{\left (6^{{1}/{3}}-\left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right ) \left (6^{{1}/{3}}+\left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right )} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(x) \,dx} \\ \int { \frac {\left (6^{{1}/{3}}-\left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right ) \left (6^{{1}/{3}}+\left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right )}{\left (-9 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} p +2 \,6^{{2}/{3}}+2 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{2}/{3}}\right ) \sqrt {p \left (4+243 p \right )}}\,dp} &= \int { \frac {\sqrt {3}}{18 x} \,dx} \\ \end{align*}

\[ \int _{}^{p \left (x \right )}\frac {\left (6^{{1}/{3}}-\left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}}\right ) \left (6^{{1}/{3}}+\left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}}\right )}{\left (-9 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} \tau +2 \,6^{{2}/{3}}+2 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\tau \left (4+243 \tau \right )}}d \tau = \sqrt {3}\, \ln \left (x^{{1}/{18}}\right )+c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(p)\) is zero, since we had to divide by this above. Solving \(g(p)=0\) or

\[ \frac {\left (-9 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} p +2 \,6^{{2}/{3}}+2 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{2}/{3}}\right ) \sqrt {p \left (4+243 p \right )}}{\left (6^{{1}/{3}}-\left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right ) \left (6^{{1}/{3}}+\left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right )}=0 \]

for \(p \left (x \right )\) gives

\begin{align*} p \left (x \right )&=-{\frac {4}{243}}\\ p \left (x \right )&=-\frac {6^{{1}/{3}} \left (4 \,6^{{1}/{3}}+\sqrt {15}\, 6^{{1}/{3}}\right )^{5}}{162}+\frac {55 \,6^{{2}/{3}} \left (4 \,6^{{1}/{3}}+\sqrt {15}\, 6^{{1}/{3}}\right )}{27}+\frac {56 \,6^{{1}/{3}} \left (4 \,6^{{1}/{3}}+\sqrt {15}\, 6^{{1}/{3}}\right )^{2}}{27}+2 \left (4 \,6^{{1}/{3}}+\sqrt {15}\, 6^{{1}/{3}}\right )^{3}+\frac {2}{9} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \int _{}^{p \left (x \right )}\frac {\left (6^{{1}/{3}}-\left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}}\right ) \left (6^{{1}/{3}}+\left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}}\right )}{\left (-9 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} \tau +2 \,6^{{2}/{3}}+2 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\tau \left (4+243 \tau \right )}}d \tau &= \sqrt {3}\, \ln \left (x^{{1}/{18}}\right )+c_1 \\ p \left (x \right ) &= -{\frac {4}{243}} \\ p \left (x \right ) &= -\frac {6^{{1}/{3}} \left (4 \,6^{{1}/{3}}+\sqrt {15}\, 6^{{1}/{3}}\right )^{5}}{162}+\frac {55 \,6^{{2}/{3}} \left (4 \,6^{{1}/{3}}+\sqrt {15}\, 6^{{1}/{3}}\right )}{27}+\frac {56 \,6^{{1}/{3}} \left (4 \,6^{{1}/{3}}+\sqrt {15}\, 6^{{1}/{3}}\right )^{2}}{27}+2 \left (4 \,6^{{1}/{3}}+\sqrt {15}\, 6^{{1}/{3}}\right )^{3}+\frac {2}{9} \\ \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} \text {Expression too large to display} \\ u \left (x \right ) &= \frac {4 x}{9} \\ u \left (x \right ) &= 2 x \\ \end{align*}

Solving ode 2A

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(u \left (x \right )=x f + g\) to (1A) shows that

\begin{align*} f &= \frac {\left (-1+i \sqrt {3}\right ) \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}}-4 i \sqrt {3}+4 \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{1}/{3}}-4}{18 \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{1}/{3}}}\\ g &= 0 \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p -\frac {\left (-1+i \sqrt {3}\right ) \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}}-4 i \sqrt {3}+4 \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{1}/{3}}-4}{18 \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{1}/{3}}} = \left (-\frac {18 x}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}}}-\frac {4 x \sqrt {3}}{3 \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}} \sqrt {243 p^{2}+4 p}}-\frac {162 x \sqrt {3}\, p}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}} \sqrt {243 p^{2}+4 p}}+\frac {16 i x}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{4}/{3}} \sqrt {243 p^{2}+4 p}}+\frac {486 i x p}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}} \sqrt {243 p^{2}+4 p}}+\frac {18 i x \sqrt {3}}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}}}+\frac {4 i x}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}} \sqrt {243 p^{2}+4 p}}+\frac {1944 i x p}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{4}/{3}} \sqrt {243 p^{2}+4 p}}+\frac {72 i x \sqrt {3}}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{4}/{3}}}+\frac {72 x}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{4}/{3}}}+\frac {16 x \sqrt {3}}{3 \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{4}/{3}} \sqrt {243 p^{2}+4 p}}+\frac {648 x \sqrt {3}\, p}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{4}/{3}} \sqrt {243 p^{2}+4 p}}\right ) p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p -\frac {\left (-1+i \sqrt {3}\right ) \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}}-4 i \sqrt {3}+4 \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{1}/{3}}-4}{18 \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{1}/{3}}} = 0 \end{align*}

No valid singular solutions found.

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )-\frac {\left (-1+i \sqrt {3}\right ) \left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{2}/{3}}-4 i \sqrt {3}+4 \left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{1}/{3}}-4}{18 \left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{1}/{3}}}}{-\frac {18 x}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{2}/{3}}}-\frac {4 x \sqrt {3}}{3 \left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{2}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}-\frac {162 x \sqrt {3}\, p \left (x \right )}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{2}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}+\frac {16 i x}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{4}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}+\frac {486 i x p \left (x \right )}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{2}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}+\frac {18 i x \sqrt {3}}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{2}/{3}}}+\frac {4 i x}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{2}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}+\frac {1944 i x p \left (x \right )}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{4}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}+\frac {72 i x \sqrt {3}}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{4}/{3}}}+\frac {72 x}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{4}/{3}}}+\frac {16 x \sqrt {3}}{3 \left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{4}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}+\frac {648 x \sqrt {3}\, p \left (x \right )}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{4}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

The ode

\begin{equation} p^{\prime }\left (x \right ) = \frac {\sqrt {3}\, \left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} p \left (x \right ) 3^{{5}/{6}} 2^{{1}/{3}}-2 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}-2 \,3^{{2}/{3}} 2^{{2}/{3}}+9 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} p \left (x \right ) 3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}+4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}}{9 \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}}\right ) x} \end{equation}

is separable as it can be written as

\begin{align*} p^{\prime }\left (x \right )&= \frac {\sqrt {3}\, \left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} p \left (x \right ) 3^{{5}/{6}} 2^{{1}/{3}}-2 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}-2 \,3^{{2}/{3}} 2^{{2}/{3}}+9 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} p \left (x \right ) 3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}+4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}}{9 \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}}\right ) x}\\ &= f(x) g(p) \end{align*}

Where

\begin{align*} f(x) &= \frac {\sqrt {3}}{9 x}\\ g(p) &= \frac {\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} p 3^{{5}/{6}} 2^{{1}/{3}}-2 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}-2 \,3^{{2}/{3}} 2^{{2}/{3}}+9 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} p 3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}+4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{2}/{3}}\right ) \sqrt {p \left (4+243 p \right )}}{\left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right )} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(x) \,dx} \\ \int { \frac {\left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right )}{\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} p 3^{{5}/{6}} 2^{{1}/{3}}-2 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}-2 \,3^{{2}/{3}} 2^{{2}/{3}}+9 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} p 3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}+4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{2}/{3}}\right ) \sqrt {p \left (4+243 p \right )}}\,dp} &= \int { \frac {\sqrt {3}}{9 x} \,dx} \\ \end{align*}

\[ \int _{}^{p \left (x \right )}\frac {\left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}}\right )}{\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} \tau 3^{{5}/{6}} 2^{{1}/{3}}-2 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}-2 \,3^{{2}/{3}} 2^{{2}/{3}}+9 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} \tau 3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}+4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\tau \left (4+243 \tau \right )}}d \tau = \sqrt {3}\, \ln \left (x^{{1}/{9}}\right )+c_2 \]

We now need to find the singular solutions, these are found by finding for what values \(g(p)\) is zero, since we had to divide by this above. Solving \(g(p)=0\) or

\[ \frac {\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} p 3^{{5}/{6}} 2^{{1}/{3}}-2 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}-2 \,3^{{2}/{3}} 2^{{2}/{3}}+9 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} p 3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}+4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{2}/{3}}\right ) \sqrt {p \left (4+243 p \right )}}{\left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right )}=0 \]

for \(p \left (x \right )\) gives

\begin{align*} p \left (x \right )&=0\\ p \left (x \right )&=-\frac {-15 i 3^{{5}/{6}} 2^{{1}/{3}}-3 i \left (\frac {7 i 3^{{5}/{6}} 2^{{1}/{3}}}{4}+\frac {7 \,3^{{1}/{3}} 2^{{1}/{3}}}{4}-\frac {3 \sqrt {-10 \,3^{{2}/{3}} 2^{{2}/{3}}+30 i 3^{{1}/{6}} 2^{{2}/{3}}}}{4}\right )^{2} 3^{{1}/{6}} 2^{{2}/{3}}+\left (\frac {7 i 3^{{5}/{6}} 2^{{1}/{3}}}{4}+\frac {7 \,3^{{1}/{3}} 2^{{1}/{3}}}{4}-\frac {3 \sqrt {-10 \,3^{{2}/{3}} 2^{{2}/{3}}+30 i 3^{{1}/{6}} 2^{{2}/{3}}}}{4}\right )^{2} 3^{{2}/{3}} 2^{{2}/{3}}-15 \,3^{{1}/{3}} 2^{{1}/{3}}+9 \sqrt {-10 \,3^{{2}/{3}} 2^{{2}/{3}}+30 i 3^{{1}/{6}} 2^{{2}/{3}}}}{54 \left (\frac {7 i 3^{{5}/{6}} 2^{{1}/{3}}}{4}+\frac {7 \,3^{{1}/{3}} 2^{{1}/{3}}}{4}-\frac {3 \sqrt {-10 \,3^{{2}/{3}} 2^{{2}/{3}}+30 i 3^{{1}/{6}} 2^{{2}/{3}}}}{4}\right )} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \int _{}^{p \left (x \right )}\frac {\left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}}\right )}{\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} \tau 3^{{5}/{6}} 2^{{1}/{3}}-2 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}-2 \,3^{{2}/{3}} 2^{{2}/{3}}+9 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} \tau 3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}+4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\tau \left (4+243 \tau \right )}}d \tau &= \sqrt {3}\, \ln \left (x^{{1}/{9}}\right )+c_2 \\ p \left (x \right ) &= 0 \\ p \left (x \right ) &= -\frac {-15 i 3^{{5}/{6}} 2^{{1}/{3}}-3 i \left (\frac {7 i 3^{{5}/{6}} 2^{{1}/{3}}}{4}+\frac {7 \,3^{{1}/{3}} 2^{{1}/{3}}}{4}-\frac {3 \sqrt {-10 \,3^{{2}/{3}} 2^{{2}/{3}}+30 i 3^{{1}/{6}} 2^{{2}/{3}}}}{4}\right )^{2} 3^{{1}/{6}} 2^{{2}/{3}}+\left (\frac {7 i 3^{{5}/{6}} 2^{{1}/{3}}}{4}+\frac {7 \,3^{{1}/{3}} 2^{{1}/{3}}}{4}-\frac {3 \sqrt {-10 \,3^{{2}/{3}} 2^{{2}/{3}}+30 i 3^{{1}/{6}} 2^{{2}/{3}}}}{4}\right )^{2} 3^{{2}/{3}} 2^{{2}/{3}}-15 \,3^{{1}/{3}} 2^{{1}/{3}}+9 \sqrt {-10 \,3^{{2}/{3}} 2^{{2}/{3}}+30 i 3^{{1}/{6}} 2^{{2}/{3}}}}{54 \left (\frac {7 i 3^{{5}/{6}} 2^{{1}/{3}}}{4}+\frac {7 \,3^{{1}/{3}} 2^{{1}/{3}}}{4}-\frac {3 \sqrt {-10 \,3^{{2}/{3}} 2^{{2}/{3}}+30 i 3^{{1}/{6}} 2^{{2}/{3}}}}{4}\right )} \\ \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} \text {Expression too large to display} \\ u \left (x \right ) &= 0 \\ u \left (x \right ) &= -\frac {4 x}{3} \\ \end{align*}

Solving ode 3A

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(u \left (x \right )=x f + g\) to (1A) shows that

\begin{align*} f &= \frac {i \left (4-\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}}\right ) \sqrt {3}-{\left (\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{1}/{3}}-2\right )}^{2}}{18 \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{1}/{3}}}\\ g &= 0 \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p -\frac {i \left (4-\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}}\right ) \sqrt {3}-{\left (\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{1}/{3}}-2\right )}^{2}}{18 \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{1}/{3}}} = \left (-\frac {16 i x}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{4}/{3}} \sqrt {243 p^{2}+4 p}}-\frac {1944 i x p}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{4}/{3}} \sqrt {243 p^{2}+4 p}}-\frac {18 i x \sqrt {3}}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}}}-\frac {18 x}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}}}-\frac {162 x \sqrt {3}\, p}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}} \sqrt {243 p^{2}+4 p}}-\frac {4 x \sqrt {3}}{3 \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}} \sqrt {243 p^{2}+4 p}}-\frac {486 i x p}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}} \sqrt {243 p^{2}+4 p}}-\frac {72 i x \sqrt {3}}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{4}/{3}}}-\frac {4 i x}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}} \sqrt {243 p^{2}+4 p}}+\frac {72 x}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{4}/{3}}}+\frac {648 x \sqrt {3}\, p}{\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{4}/{3}} \sqrt {243 p^{2}+4 p}}+\frac {16 x \sqrt {3}}{3 \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{4}/{3}} \sqrt {243 p^{2}+4 p}}\right ) p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p -\frac {i \left (4-\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{2}/{3}}\right ) \sqrt {3}-{\left (\left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{1}/{3}}-2\right )}^{2}}{18 \left (8+972 p +36 \sqrt {3}\, \sqrt {243 p^{2}+4 p}\right )^{{1}/{3}}} = 0 \end{align*}

No valid singular solutions found.

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )-\frac {i \left (4-\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{2}/{3}}\right ) \sqrt {3}-{\left (\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{1}/{3}}-2\right )}^{2}}{18 \left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{1}/{3}}}}{-\frac {16 i x}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{4}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}-\frac {1944 i x p \left (x \right )}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{4}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}-\frac {18 i x \sqrt {3}}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{2}/{3}}}-\frac {18 x}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{2}/{3}}}-\frac {162 x \sqrt {3}\, p \left (x \right )}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{2}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}-\frac {4 x \sqrt {3}}{3 \left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{2}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}-\frac {486 i x p \left (x \right )}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{2}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}-\frac {72 i x \sqrt {3}}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{4}/{3}}}-\frac {4 i x}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{2}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}+\frac {72 x}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{4}/{3}}}+\frac {648 x \sqrt {3}\, p \left (x \right )}{\left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{4}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}+\frac {16 x \sqrt {3}}{3 \left (8+972 p \left (x \right )+36 \sqrt {3}\, \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}\right )^{{4}/{3}} \sqrt {243 p \left (x \right )^{2}+4 p \left (x \right )}}} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

The ode

\begin{equation} p^{\prime }\left (x \right ) = \frac {\sqrt {3}\, \left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} p \left (x \right ) 3^{{5}/{6}} 2^{{1}/{3}}-2 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}+2 \,3^{{2}/{3}} 2^{{2}/{3}}-9 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} p \left (x \right ) 3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}-4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}}{9 \left (i 3^{{5}/{6}} 2^{{1}/{3}}-3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}}\right ) \left (-i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}}\right ) x} \end{equation}

is separable as it can be written as

\begin{align*} p^{\prime }\left (x \right )&= \frac {\sqrt {3}\, \left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} p \left (x \right ) 3^{{5}/{6}} 2^{{1}/{3}}-2 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}+2 \,3^{{2}/{3}} 2^{{2}/{3}}-9 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} p \left (x \right ) 3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}-4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}}{9 \left (i 3^{{5}/{6}} 2^{{1}/{3}}-3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}}\right ) \left (-i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+2 \sqrt {3}+27 \sqrt {p \left (x \right ) \left (4+243 p \left (x \right )\right )}\right )\right )^{{1}/{3}}\right ) x}\\ &= f(x) g(p) \end{align*}

Where

\begin{align*} f(x) &= \frac {\sqrt {3}}{9 x}\\ g(p) &= \frac {\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} p 3^{{5}/{6}} 2^{{1}/{3}}-2 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}+2 \,3^{{2}/{3}} 2^{{2}/{3}}-9 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} p 3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}-4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{2}/{3}}\right ) \sqrt {p \left (4+243 p \right )}}{\left (i 3^{{5}/{6}} 2^{{1}/{3}}-3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right ) \left (-i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right )} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(x) \,dx} \\ \int { \frac {\left (i 3^{{5}/{6}} 2^{{1}/{3}}-3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right ) \left (-i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right )}{\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} p 3^{{5}/{6}} 2^{{1}/{3}}-2 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}+2 \,3^{{2}/{3}} 2^{{2}/{3}}-9 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} p 3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}-4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{2}/{3}}\right ) \sqrt {p \left (4+243 p \right )}}\,dp} &= \int { \frac {\sqrt {3}}{9 x} \,dx} \\ \end{align*}

\[ \int _{}^{p \left (x \right )}\frac {\left (i 3^{{5}/{6}} 2^{{1}/{3}}-3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}}\right ) \left (-i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}}\right )}{\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} \tau 3^{{5}/{6}} 2^{{1}/{3}}-2 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}+2 \,3^{{2}/{3}} 2^{{2}/{3}}-9 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} \tau 3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}-4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\tau \left (4+243 \tau \right )}}d \tau = \sqrt {3}\, \ln \left (x^{{1}/{9}}\right )+c_3 \]

We now need to find the singular solutions, these are found by finding for what values \(g(p)\) is zero, since we had to divide by this above. Solving \(g(p)=0\) or

\[ \frac {\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} p 3^{{5}/{6}} 2^{{1}/{3}}-2 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}+2 \,3^{{2}/{3}} 2^{{2}/{3}}-9 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} p 3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}-4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{2}/{3}}\right ) \sqrt {p \left (4+243 p \right )}}{\left (i 3^{{5}/{6}} 2^{{1}/{3}}-3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right ) \left (-i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +2 \sqrt {3}+27 \sqrt {p \left (4+243 p \right )}\right )\right )^{{1}/{3}}\right )}=0 \]

for \(p \left (x \right )\) gives

\begin{align*} p \left (x \right )&=0\\ p \left (x \right )&=-{\frac {4}{243}} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \int _{}^{p \left (x \right )}\frac {\left (i 3^{{5}/{6}} 2^{{1}/{3}}-3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}}\right ) \left (-i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}}\right )}{\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} \tau 3^{{5}/{6}} 2^{{1}/{3}}-2 i \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}+2 \,3^{{2}/{3}} 2^{{2}/{3}}-9 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} \tau 3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}-4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +2 \sqrt {3}+27 \sqrt {\tau \left (4+243 \tau \right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\tau \left (4+243 \tau \right )}}d \tau &= \sqrt {3}\, \ln \left (x^{{1}/{9}}\right )+c_3 \\ p \left (x \right ) &= 0 \\ p \left (x \right ) &= -{\frac {4}{243}} \\ \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} \text {Expression too large to display} \\ u \left (x \right ) &= 0 \\ u \left (x \right ) &= \frac {4 x}{9} \\ \end{align*}

The solution \(\text {Expression too large to display}\) simplifies to

\begin{align*} \text {Expression too large to display} \\ \end{align*}

The solution \(\text {Expression too large to display}\) simplifies to

\begin{align*} \text {Expression too large to display} \\ \end{align*}

The solution \(\text {Expression too large to display}\) simplifies to

\begin{align*} \text {Expression too large to display} \\ \end{align*}

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

\begin{align*} u \left (x \right ) &= 0 \\ u \left (x \right ) &= 2 x \\ u \left (x \right ) &= -\frac {4 x}{3} \\ \end{align*}

2.1.16.9 Solved using first_order_ode_homog_type_D2

2.067 (sec)

Entering first order ode homog type D2 solver

\begin{align*} u^{\prime }\left (x \right )&=\frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )^{2}}{4 x^{2}} \\ \end{align*}

Applying change of variables \(u \left (x \right ) = u \left (x \right ) x\), then the ode becomes

\begin{align*} u^{\prime }\left (x \right ) x +u \left (x \right ) = \frac {3 u \left (x \right )^{3}}{8}-\frac {u \left (x \right )^{2}}{4} \end{align*}

Which is now solved The ode

\begin{equation} u^{\prime }\left (x \right ) = \frac {u \left (x \right ) \left (3 u \left (x \right )+4\right ) \left (u \left (x \right )-2\right )}{8 x} \end{equation}

is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= \frac {u \left (x \right ) \left (3 u \left (x \right )+4\right ) \left (u \left (x \right )-2\right )}{8 x}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{8 x}\\ g(u) &= u \left (3 u +4\right ) \left (u -2\right ) \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\ \int { \frac {1}{u \left (3 u +4\right ) \left (u -2\right )}\,du} &= \int { \frac {1}{8 x} \,dx} \\ \end{align*}

\[ -\frac {\ln \left (u \left (x \right )\right )}{8}+\frac {\ln \left (u \left (x \right )-2\right )}{20}+\frac {3 \ln \left (3 u \left (x \right )+4\right )}{40}=\ln \left (x^{{1}/{8}}\right )+c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ u \left (3 u +4\right ) \left (u -2\right )=0 \]

for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=0\\ u \left (x \right )&=2 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} -\frac {\ln \left (u \left (x \right )\right )}{8}+\frac {\ln \left (u \left (x \right )-2\right )}{20}+\frac {3 \ln \left (3 u \left (x \right )+4\right )}{40} &= \ln \left (x^{{1}/{8}}\right )+c_1 \\ u \left (x \right ) &= 0 \\ u \left (x \right ) &= 2 \\ \end{align*}

Converting \(-\frac {\ln \left (u \left (x \right )\right )}{8}+\frac {\ln \left (u \left (x \right )-2\right )}{20}+\frac {3 \ln \left (3 u \left (x \right )+4\right )}{40} = \ln \left (x^{{1}/{8}}\right )+c_1\) back to \(u \left (x \right )\) gives

\begin{align*} -\frac {\ln \left (\frac {u \left (x \right )}{x}\right )}{8}+\frac {\ln \left (\frac {u \left (x \right )}{x}-2\right )}{20}+\frac {3 \ln \left (\frac {3 u \left (x \right )}{x}+4\right )}{40} = \ln \left (x^{{1}/{8}}\right )+c_1 \end{align*}

Converting \(u \left (x \right ) = 0\) back to \(u \left (x \right )\) gives

\begin{align*} u \left (x \right ) = 0 \end{align*}

Converting \(u \left (x \right ) = 2\) back to \(u \left (x \right )\) gives

\begin{align*} u \left (x \right ) = 2 x \end{align*}

Simplifying the above gives

\begin{align*} -\frac {\ln \left (\frac {u \left (x \right )}{x}\right )}{8}+\frac {\ln \left (\frac {u \left (x \right )-2 x}{x}\right )}{20}+\frac {3 \ln \left (\frac {3 u \left (x \right )+4 x}{x}\right )}{40} &= \frac {\ln \left (x \right )}{8}+c_1 \\ u \left (x \right ) &= 0 \\ u \left (x \right ) &= 2 x \\ \end{align*}

Summary of solutions found

\begin{align*} -\frac {\ln \left (\frac {u \left (x \right )}{x}\right )}{8}+\frac {\ln \left (\frac {u \left (x \right )-2 x}{x}\right )}{20}+\frac {3 \ln \left (\frac {3 u \left (x \right )+4 x}{x}\right )}{40} &= \frac {\ln \left (x \right )}{8}+c_1 \\ u \left (x \right ) &= 0 \\ u \left (x \right ) &= 2 x \\ \end{align*}

2.1.16.10 Solved using first_order_ode_homog_type_G

1.913 (sec)

Entering first order ode homog type G solver

\begin{align*} u^{\prime }\left (x \right )&=\frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )^{2}}{4 x^{2}} \\ \end{align*}

Multiplying the right side of the ode, which is \(\frac {3 u^{3}}{8 x^{3}}-\frac {u^{2}}{4 x^{2}}\) by \(\frac {x}{u}\) gives

\begin{align*} u^{\prime }\left (x \right ) &= \left (\frac {x}{u}\right ) \frac {3 u^{3}}{8 x^{3}}-\frac {u^{2}}{4 x^{2}}\\ &= \frac {u \left (3 u -2 x \right )}{8 x^{2}}\\ &= F(x,u) \end{align*}

Since \(F \left (x , u\right )\) has \(u\), then let

\begin{align*} f_x&= x \left (\frac {\partial }{\partial x}F \left (x , u\right )\right )\\ &= -\frac {u \left (3 u -x \right )}{4 x^{2}}\\ f_u&= u \left (\frac {\partial }{\partial u}F \left (x , u\right )\right )\\ &= \frac {u \left (3 u -x \right )}{4 x^{2}}\\ \alpha &= \frac {f_x}{f_u} \\ &=-1 \end{align*}

Since \(\alpha \) is independent of \(x,u\) then this is Homogeneous type G.

Let

\begin{align*} u&=\frac {z}{x^ \alpha }\\ &=\frac {z}{\frac {1}{x}} \end{align*}

Substituting the above back into \(F(x,u)\) gives

\begin{align*} F \left (z \right ) &=\frac {3}{8} z^{2}-\frac {1}{4} z \end{align*}

We see that \(F \left (z \right )\) does not depend on \(x\) nor on \(u\). If this was not the case, then this method will not work.

Therefore, the implicit solution is given by

\begin{align*} \ln \left (x \right )- c_1 - \int ^{u x^\alpha } \frac {1}{z \left (\alpha + F(z)\right ) } \,dz & = 0 \end{align*}

Which gives

\[ \ln \left (x \right )-c_1 +\int _{}^{\frac {u \left (x \right )}{x}}\frac {1}{z \left (1-\frac {3}{8} z^{2}+\frac {1}{4} z \right )}d z = 0 \]

The value of the above is

\[ \ln \left (x \right )-c_1 +\ln \left (\frac {u \left (x \right )}{x}\right )-\frac {2 \ln \left (\frac {u \left (x \right )-2 x}{x}\right )}{5}-\frac {3 \ln \left (\frac {3 u \left (x \right )+4 x}{x}\right )}{5} = 0 \]

Summary of solutions found

\begin{align*} \ln \left (x \right )-c_1 +\ln \left (\frac {u \left (x \right )}{x}\right )-\frac {2 \ln \left (\frac {u \left (x \right )-2 x}{x}\right )}{5}-\frac {3 \ln \left (\frac {3 u \left (x \right )+4 x}{x}\right )}{5} &= 0 \\ \end{align*}

2.1.16.11 Solved using first_order_ode_homog_type_maple_C

12.819 (sec)

Entering first order ode homog type maple C solver

\begin{align*} u^{\prime }\left (x \right )&=\frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )^{2}}{4 x^{2}} \\ \end{align*}

Let \(Y = u -y_{0}\) and \(X = x -x_{0}\) then the above is transformed to new ode in \(Y(X)\)

\[ \frac {d}{d X}Y \left (X \right ) = \frac {\left (Y \left (X \right )+y_{0} \right )^{2} \left (3 Y \left (X \right )+3 y_{0} -2 X -2 x_{0} \right )}{8 \left (X +x_{0} \right )^{3}} \]

Solving for possible values of \(x_{0}\) and \(y_{0}\) which makes the above ode a homogeneous ode results in

\begin{align*} x_{0}&=0\\ y_{0}&=0 \end{align*}

Using these values now it is possible to easily solve for \(Y \left (X \right )\). The above ode now becomes

\begin{align*} \frac {d}{d X}Y \left (X \right ) = \frac {-2 Y \left (X \right )^{2} X +3 Y \left (X \right )^{3}}{8 X^{3}} \end{align*}

In canonical form, the ODE is

\begin{align*} Y' &= F(X,Y)\\ &= \frac {-2 Y^{2} X +3 Y^{3}}{8 X^{3}}\tag {1} \end{align*}

An ode of the form \(Y' = \frac {M(X,Y)}{N(X,Y)}\) is called homogeneous if the functions \(M(X,Y)\) and \(N(X,Y)\) are both homogeneous functions and of the same order. Recall that a function \(f(X,Y)\) is homogeneous of order \(n\) if

\[ f(t^n X, t^n Y)= t^n f(X,Y) \]

In this case, it can be seen that both \(M=-Y^{2} \left (2 X -3 Y \right )\) and \(N=8 X^{3}\) are both homogeneous and of the same order \(n=3\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {Y}{X}\), or \(Y=uX\). Hence

\[ \frac { \mathop {\mathrm {d}Y}}{\mathop {\mathrm {d}X}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}}X + u \]

Applying the transformation \(Y=uX\) to the above ODE in (1) gives

\begin{align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}}X + u &= -\frac {1}{4} u^{2}+\frac {3}{8} u^{3}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {-\frac {u \left (X \right )^{2}}{4}+\frac {3 u \left (X \right )^{3}}{8}-u \left (X \right )}{X} \end{align*}

Or

\[ \frac {d}{d X}u \left (X \right )-\frac {-\frac {u \left (X \right )^{2}}{4}+\frac {3 u \left (X \right )^{3}}{8}-u \left (X \right )}{X} = 0 \]

Or

\[ -3 u \left (X \right )^{3}+2 u \left (X \right )^{2}+8 \left (\frac {d}{d X}u \left (X \right )\right ) X +8 u \left (X \right ) = 0 \]

Which is now solved as separable in \(u \left (X \right )\).

The ode

\begin{equation} \frac {d}{d X}u \left (X \right ) = \frac {u \left (X \right ) \left (3 u \left (X \right )+4\right ) \left (u \left (X \right )-2\right )}{8 X} \end{equation}

is separable as it can be written as

\begin{align*} \frac {d}{d X}u \left (X \right )&= \frac {u \left (X \right ) \left (3 u \left (X \right )+4\right ) \left (u \left (X \right )-2\right )}{8 X}\\ &= f(X) g(u) \end{align*}

Where

\begin{align*} f(X) &= \frac {1}{8 X}\\ g(u) &= u \left (3 u +4\right ) \left (u -2\right ) \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(X) \,dX} \\ \int { \frac {1}{u \left (3 u +4\right ) \left (u -2\right )}\,du} &= \int { \frac {1}{8 X} \,dX} \\ \end{align*}

\[ -\frac {\ln \left (u \left (X \right )\right )}{8}+\frac {\ln \left (u \left (X \right )-2\right )}{20}+\frac {3 \ln \left (3 u \left (X \right )+4\right )}{40}=\ln \left (X^{{1}/{8}}\right )+c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ u \left (3 u +4\right ) \left (u -2\right )=0 \]

for \(u \left (X \right )\) gives

\begin{align*} u \left (X \right )&=0\\ u \left (X \right )&=2 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} -\frac {\ln \left (u \left (X \right )\right )}{8}+\frac {\ln \left (u \left (X \right )-2\right )}{20}+\frac {3 \ln \left (3 u \left (X \right )+4\right )}{40} &= \ln \left (X^{{1}/{8}}\right )+c_1 \\ u \left (X \right ) &= 0 \\ u \left (X \right ) &= 2 \\ \end{align*}

Converting \(-\frac {\ln \left (u \left (X \right )\right )}{8}+\frac {\ln \left (u \left (X \right )-2\right )}{20}+\frac {3 \ln \left (3 u \left (X \right )+4\right )}{40} = \ln \left (X^{{1}/{8}}\right )+c_1\) back to \(Y \left (X \right )\) gives

\begin{align*} -\ln \left (\frac {Y \left (X \right )}{X}\right )+\frac {2 \ln \left (\frac {Y \left (X \right )-2 X}{X}\right )}{5}+\frac {3 \ln \left (\frac {3 Y \left (X \right )+4 X}{X}\right )}{5} = \ln \left (X \right )+8 c_1 \end{align*}

Converting \(u \left (X \right ) = 0\) back to \(Y \left (X \right )\) gives

\begin{align*} Y \left (X \right ) = 0 \end{align*}

Converting \(u \left (X \right ) = 2\) back to \(Y \left (X \right )\) gives

\begin{align*} Y \left (X \right ) = 2 X \end{align*}

Using the solution for \(Y(X)\)

\begin{align*} -\ln \left (\frac {Y \left (X \right )}{X}\right )+\frac {2 \ln \left (\frac {Y \left (X \right )-2 X}{X}\right )}{5}+\frac {3 \ln \left (\frac {3 Y \left (X \right )+4 X}{X}\right )}{5} = \ln \left (X \right )+8 c_1\tag {A} \end{align*}

And replacing back terms in the above solution using

\begin{align*} Y &= u +y_{0}\\ X &= x +x_{0} \end{align*}

Or

\begin{align*} Y &= u\\ X &= x \end{align*}

Then the solution in \(u \left (x \right )\) becomes using EQ (A)

\begin{align*} -\ln \left (\frac {u \left (x \right )}{x}\right )+\frac {2 \ln \left (\frac {u \left (x \right )-2 x}{x}\right )}{5}+\frac {3 \ln \left (\frac {3 u \left (x \right )+4 x}{x}\right )}{5} = \ln \left (x \right )+8 c_1 \end{align*}

Using the solution for \(Y(X)\)

\begin{align*} Y \left (X \right ) = 0\tag {A} \end{align*}

And replacing back terms in the above solution using

\begin{align*} Y &= u +y_{0}\\ X &= x +x_{0} \end{align*}

Or

\begin{align*} Y &= u\\ X &= x \end{align*}

Then the solution in \(u \left (x \right )\) becomes using EQ (A)

\begin{align*} u \left (x \right ) = 0 \end{align*}

Using the solution for \(Y(X)\)

\begin{align*} Y \left (X \right ) = 2 X\tag {A} \end{align*}

And replacing back terms in the above solution using

\begin{align*} Y &= u +y_{0}\\ X &= x +x_{0} \end{align*}

Or

\begin{align*} Y &= u\\ X &= x \end{align*}

Then the solution in \(u \left (x \right )\) becomes using EQ (A)

\begin{align*} u \left (x \right ) = 2 x \end{align*}

Summary of solutions found

\begin{align*} -\ln \left (\frac {u \left (x \right )}{x}\right )+\frac {2 \ln \left (\frac {u \left (x \right )-2 x}{x}\right )}{5}+\frac {3 \ln \left (\frac {3 u \left (x \right )+4 x}{x}\right )}{5} &= \ln \left (x \right )+8 c_1 \\ u \left (x \right ) &= 0 \\ u \left (x \right ) &= 2 x \\ \end{align*}

2.1.16.12 Solved using first_order_ode_abel_first_kind

110.922 (sec)

Entering first order ode abel first kind solver

\begin{align*} u^{\prime }\left (x \right )&=\frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )^{2}}{4 x^{2}} \\ \end{align*}

This is Abel first kind ODE, it has the form

\[ u^{\prime }\left (x \right )= f_0(x)+f_1(x) u \left (x \right ) +f_2(x)u \left (x \right )^{2}+f_3(x)u \left (x \right )^{3} \]

Comparing the above to given ODE which is

\begin{align*}u^{\prime }\left (x \right )&=\frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )^{2}}{4 x^{2}}\tag {1} \end{align*}

Therefore

\begin{align*} f_0 &= 0\\ f_1 &= 0\\ f_2 &= -\frac {1}{4 x^{2}}\\ f_3 &= \frac {3}{8 x^{3}} \end{align*}

Hence

\begin{align*} f'_{0} &= 0\\ f'_{3} &= -\frac {9}{8 x^{4}} \end{align*}

Since \(f_2(x)=-\frac {1}{4 x^{2}}\) is not zero, then the followingtransformation is used to remove \(f_2\). Let \(u \left (x \right ) = u(x) - \frac {f_2}{3 f_3}\) or

\begin{align*} u \left (x \right ) &= u(x) - \left ( \frac {-\frac {1}{4 x^{2}}}{\frac {9}{8 x^{3}}} \right ) \\ &= u \left (x \right )+\frac {2 x}{9} \end{align*}

The above transformation applied to (1) gives a new ODE as

\begin{align*} u^{\prime }\left (x \right ) = \frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )}{18 x}-\frac {56}{243}\tag {2} \end{align*}

The above ODE (2) can now be solved.

2.1.16.13 Solved using first_order_ode_isobaric

2.163 (sec)

Entering first order ode isobaric solverSolving for \(u'\) gives

\[ u' = \frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )}{18 x}-\frac {56}{243} \]

An ode \(u^{\prime }\left (x \right )=f(x,u)\) is isobaric if

\[ f(t x, t^m u) = t^{m-1} f(x,u)\tag {1} \]

Where here

\[ f(x,u) = \frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )}{18 x}-\frac {56}{243}\tag {2} \]

\(m\) is the order of isobaric. Substituting (2) into (1) and solving for \(m\) gives

\[ m = 1 \]

Since the ode is isobaric of order \(m=1\), then the substitution

\begin{align*} u&=u x^m \\ &=u x \end{align*}

Converts the ODE to a separable in \(u \left (x \right )\). Performing this substitution gives

\[ u \left (x \right )+x u^{\prime }\left (x \right ) = \frac {3 u \left (x \right )^{3}}{8}-\frac {u \left (x \right )}{18}-\frac {56}{243} \]

Entering first order ode separable solverThe ode

\begin{equation} u^{\prime }\left (x \right ) = \frac {\left (9 u \left (x \right )+2\right ) \left (9 u \left (x \right )-16\right ) \left (9 u \left (x \right )+14\right )}{1944 x} \end{equation}

is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= \frac {\left (9 u \left (x \right )+2\right ) \left (9 u \left (x \right )-16\right ) \left (9 u \left (x \right )+14\right )}{1944 x}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{1944 x}\\ g(u) &= \left (9 u +2\right ) \left (9 u -16\right ) \left (9 u +14\right ) \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\ \int { \frac {1}{\left (9 u +2\right ) \left (9 u -16\right ) \left (9 u +14\right )}\,du} &= \int { \frac {1}{1944 x} \,dx} \\ \end{align*}

\[ \frac {\ln \left (9 u \left (x \right )-16\right )}{4860}+\frac {\ln \left (9 u \left (x \right )+14\right )}{3240}-\frac {\ln \left (9 u \left (x \right )+2\right )}{1944}=\ln \left (x^{\frac {1}{1944}}\right )+c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ \left (9 u +2\right ) \left (9 u -16\right ) \left (9 u +14\right )=0 \]

for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=-{\frac {14}{9}}\\ u \left (x \right )&=-{\frac {2}{9}} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \frac {\ln \left (9 u \left (x \right )-16\right )}{4860}+\frac {\ln \left (9 u \left (x \right )+14\right )}{3240}-\frac {\ln \left (9 u \left (x \right )+2\right )}{1944} &= \ln \left (x^{\frac {1}{1944}}\right )+c_1 \\ u \left (x \right ) &= -{\frac {14}{9}} \\ u \left (x \right ) &= -{\frac {2}{9}} \\ \end{align*}

Converting \(\frac {\ln \left (9 u \left (x \right )-16\right )}{4860}+\frac {\ln \left (9 u \left (x \right )+14\right )}{3240}-\frac {\ln \left (9 u \left (x \right )+2\right )}{1944} = \frac {\ln \left (x \right )}{1944}+c_1\) back to \(u \left (x \right )\) gives

\begin{align*} \frac {\ln \left (\frac {9 u \left (x \right )}{x}-16\right )}{4860}+\frac {\ln \left (\frac {9 u \left (x \right )}{x}+14\right )}{3240}-\frac {\ln \left (\frac {9 u \left (x \right )}{x}+2\right )}{1944} = \frac {\ln \left (x \right )}{1944}+c_1 \end{align*}

Converting \(u \left (x \right ) = -{\frac {14}{9}}\) back to \(u \left (x \right )\) gives

\begin{align*} \frac {u \left (x \right )}{x} = -{\frac {14}{9}} \end{align*}

Converting \(u \left (x \right ) = -{\frac {2}{9}}\) back to \(u \left (x \right )\) gives

\begin{align*} \frac {u \left (x \right )}{x} = -{\frac {2}{9}} \end{align*}

Simplifying the above gives

\begin{align*} \frac {\ln \left (\frac {9 u \left (x \right )-16 x}{x}\right )}{4860}+\frac {\ln \left (\frac {9 u \left (x \right )+14 x}{x}\right )}{3240}-\frac {\ln \left (\frac {9 u \left (x \right )+2 x}{x}\right )}{1944} &= \frac {\ln \left (x \right )}{1944}+c_1 \\ \frac {u \left (x \right )}{x} &= -{\frac {14}{9}} \\ \frac {u \left (x \right )}{x} &= -{\frac {2}{9}} \\ \end{align*}

Solving for \(u \left (x \right )\) gives

\begin{align*} \frac {\ln \left (\frac {9 u \left (x \right )-16 x}{x}\right )}{4860}+\frac {\ln \left (\frac {9 u \left (x \right )+14 x}{x}\right )}{3240}-\frac {\ln \left (\frac {9 u \left (x \right )+2 x}{x}\right )}{1944} &= \frac {\ln \left (x \right )}{1944}+c_1 \\ u \left (x \right ) &= -\frac {14 x}{9} \\ u \left (x \right ) &= -\frac {2 x}{9} \\ \end{align*}

Summary of solutions found

\begin{align*} \frac {\ln \left (\frac {9 u \left (x \right )-16 x}{x}\right )}{4860}+\frac {\ln \left (\frac {9 u \left (x \right )+14 x}{x}\right )}{3240}-\frac {\ln \left (\frac {9 u \left (x \right )+2 x}{x}\right )}{1944} &= \frac {\ln \left (x \right )}{1944}+c_1 \\ u \left (x \right ) &= -\frac {14 x}{9} \\ u \left (x \right ) &= -\frac {2 x}{9} \\ \end{align*}

2.1.16.14 Solved using first_order_ode_homog_A

2.964 (sec)

Entering first order ode homog A solver

\begin{align*} u^{\prime }\left (x \right )&=\frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )}{18 x}-\frac {56}{243} \\ \end{align*}

In canonical form, the ODE is

\begin{align*} u' &= F(x,u)\\ &= \frac {3 u^{3}}{8 x^{3}}-\frac {u}{18 x}-\frac {56}{243}\tag {1} \end{align*}

An ode of the form \(u' = \frac {M(x,u)}{N(x,u)}\) is called homogeneous if the functions \(M(x,u)\) and \(N(x,u)\) are both homogeneous functions and of the same order. Recall that a function \(f(x,u)\) is homogeneous of order \(n\) if

\[ f(t^n x, t^n u)= t^n f(x,u) \]

In this case, it can be seen that both \(M=729 u^{3}-108 u \,x^{2}-448 x^{3}\) and \(N=1944 x^{3}\) are both homogeneous and of the same order \(n=3\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {u}{x}\), or \(u=ux\). Hence

\[ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u \]

Applying the transformation \(u=ux\) to the above ODE in (1) gives

\begin{align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u &= \frac {3}{8} u^{3}-\frac {1}{18} u -\frac {56}{243}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {\frac {3 u \left (x \right )^{3}}{8}-\frac {19 u \left (x \right )}{18}-\frac {56}{243}}{x} \end{align*}

Or

\[ u^{\prime }\left (x \right )-\frac {\frac {3 u \left (x \right )^{3}}{8}-\frac {19 u \left (x \right )}{18}-\frac {56}{243}}{x} = 0 \]

Or

\[ -729 u \left (x \right )^{3}+1944 u^{\prime }\left (x \right ) x +2052 u \left (x \right )+448 = 0 \]

Which is now solved as separable in \(u \left (x \right )\).

The ode

\begin{equation} u^{\prime }\left (x \right ) = \frac {\left (9 u \left (x \right )+2\right ) \left (9 u \left (x \right )-16\right ) \left (9 u \left (x \right )+14\right )}{1944 x} \end{equation}

is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= \frac {\left (9 u \left (x \right )+2\right ) \left (9 u \left (x \right )-16\right ) \left (9 u \left (x \right )+14\right )}{1944 x}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{1944 x}\\ g(u) &= \left (9 u +2\right ) \left (9 u -16\right ) \left (9 u +14\right ) \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\ \int { \frac {1}{\left (9 u +2\right ) \left (9 u -16\right ) \left (9 u +14\right )}\,du} &= \int { \frac {1}{1944 x} \,dx} \\ \end{align*}

\[ \frac {\ln \left (9 u \left (x \right )-16\right )}{4860}+\frac {\ln \left (9 u \left (x \right )+14\right )}{3240}-\frac {\ln \left (9 u \left (x \right )+2\right )}{1944}=\ln \left (x^{\frac {1}{1944}}\right )+c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ \left (9 u +2\right ) \left (9 u -16\right ) \left (9 u +14\right )=0 \]

for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=-{\frac {14}{9}}\\ u \left (x \right )&=-{\frac {2}{9}} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \frac {\ln \left (9 u \left (x \right )-16\right )}{4860}+\frac {\ln \left (9 u \left (x \right )+14\right )}{3240}-\frac {\ln \left (9 u \left (x \right )+2\right )}{1944} &= \ln \left (x^{\frac {1}{1944}}\right )+c_1 \\ u \left (x \right ) &= -{\frac {14}{9}} \\ u \left (x \right ) &= -{\frac {2}{9}} \\ \end{align*}

Converting \(\frac {\ln \left (9 u \left (x \right )-16\right )}{4860}+\frac {\ln \left (9 u \left (x \right )+14\right )}{3240}-\frac {\ln \left (9 u \left (x \right )+2\right )}{1944} = \ln \left (x^{\frac {1}{1944}}\right )+c_1\) back to \(u \left (x \right )\) gives

\begin{align*} \frac {2 \ln \left (\frac {9 u \left (x \right )-16 x}{x}\right )}{5}+\frac {3 \ln \left (\frac {9 u \left (x \right )+14 x}{x}\right )}{5}-\ln \left (\frac {9 u \left (x \right )+2 x}{x}\right ) = \ln \left (x \right )+1944 c_1 \end{align*}

Converting \(u \left (x \right ) = -{\frac {14}{9}}\) back to \(u \left (x \right )\) gives

\begin{align*} u \left (x \right ) = -\frac {14 x}{9} \end{align*}

Converting \(u \left (x \right ) = -{\frac {2}{9}}\) back to \(u \left (x \right )\) gives

\begin{align*} u \left (x \right ) = -\frac {2 x}{9} \end{align*}

Summary of solutions found

\begin{align*} \frac {2 \ln \left (\frac {9 u \left (x \right )-16 x}{x}\right )}{5}+\frac {3 \ln \left (\frac {9 u \left (x \right )+14 x}{x}\right )}{5}-\ln \left (\frac {9 u \left (x \right )+2 x}{x}\right ) &= \ln \left (x \right )+1944 c_1 \\ u \left (x \right ) &= -\frac {14 x}{9} \\ u \left (x \right ) &= -\frac {2 x}{9} \\ \end{align*}

2.1.16.15 Solved using first_order_ode_dAlembert

76.884 (sec)

Entering first order ode dAlembert solver

\begin{align*} u^{\prime }\left (x \right )&=\frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )}{18 x}-\frac {56}{243} \\ \end{align*}

Let \(p=u^{\prime }\left (x \right )\) the ode becomes

\begin{align*} p = \frac {3 u^{3}}{8 x^{3}}-\frac {u}{18 x}-\frac {56}{243} \end{align*}

Solving for \(u \left (x \right )\) from the above results in

\begin{align*} \tag{1} u \left (x \right ) &= \left (\frac {\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}}{9}+\frac {4}{9 \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}}\right ) x \\ \tag{2} u \left (x \right ) &= \left (-\frac {\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}}{18}-\frac {2}{9 \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}}+\frac {i \sqrt {3}\, \left (\frac {\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}}{2}-\frac {2}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}}\right )}{9}\right ) x \\ \tag{3} u \left (x \right ) &= \left (-\frac {\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}}{18}-\frac {2}{9 \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}}-\frac {i \sqrt {3}\, \left (\frac {\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}}{2}-\frac {2}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}}\right )}{9}\right ) x \\ \end{align*}

This has the form

\begin{align*} u=x f(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=u'(x)\). Each of the above ode’s is dAlembert ode which is now solved.

Solving ode 1A

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(u \left (x \right )=x f + g\) to (1A) shows that

\begin{align*} f &= \frac {\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}}+4}{9 \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}}\\ g &= 0 \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p -\frac {\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}}+4}{9 \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}} = \left (\frac {36 x}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}}}+\frac {2916 x p}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}} \sqrt {6561 p^{2}+3024 p +348}}+\frac {672 x}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}} \sqrt {6561 p^{2}+3024 p +348}}-\frac {144 x}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{4}/{3}}}-\frac {11664 x p}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{4}/{3}} \sqrt {6561 p^{2}+3024 p +348}}-\frac {2688 x}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{4}/{3}} \sqrt {6561 p^{2}+3024 p +348}}\right ) p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p -\frac {\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}}+4}{9 \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}} = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &={\frac {16}{9}} \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} u \left (x \right ) = \frac {16 x}{9} \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )-\frac {\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}}+4}{9 \left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{1}/{3}}}}{\frac {36 x}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}}}+\frac {2916 x p \left (x \right )}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}+\frac {672 x}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}-\frac {144 x}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{4}/{3}}}-\frac {11664 x p \left (x \right )}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{4}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}-\frac {2688 x}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{4}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

The ode

\begin{equation} p^{\prime }\left (x \right ) = -\frac {\sqrt {3}\, \left (9 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}} p \left (x \right )-2 \,6^{{2}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}}{54 \left (6^{{1}/{3}}+\left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}}\right ) \left (6^{{1}/{3}}-\left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}}\right ) x} \end{equation}

is separable as it can be written as

\begin{align*} p^{\prime }\left (x \right )&= -\frac {\sqrt {3}\, \left (9 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}} p \left (x \right )-2 \,6^{{2}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}}{54 \left (6^{{1}/{3}}+\left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}}\right ) \left (6^{{1}/{3}}-\left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}}\right ) x}\\ &= f(x) g(p) \end{align*}

Where

\begin{align*} f(x) &= -\frac {\sqrt {3}}{54 x}\\ g(p) &= \frac {\left (9 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}} p -2 \,6^{{2}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}}{\left (6^{{1}/{3}}+\left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right ) \left (6^{{1}/{3}}-\left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right )} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(x) \,dx} \\ \int { \frac {\left (6^{{1}/{3}}+\left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right ) \left (6^{{1}/{3}}-\left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right )}{\left (9 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}} p -2 \,6^{{2}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}}\,dp} &= \int { -\frac {\sqrt {3}}{54 x} \,dx} \\ \end{align*}

\[ \int _{}^{p \left (x \right )}\frac {\left (6^{{1}/{3}}+\left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}}\right ) \left (6^{{1}/{3}}-\left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}}\right )}{\left (9 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}} \tau -2 \,6^{{2}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}}d \tau = \sqrt {3}\, \ln \left (\frac {1}{x^{{1}/{54}}}\right )+c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(p)\) is zero, since we had to divide by this above. Solving \(g(p)=0\) or

\[ \frac {\left (9 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}} p -2 \,6^{{2}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}}{\left (6^{{1}/{3}}+\left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right ) \left (6^{{1}/{3}}-\left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right )}=0 \]

for \(p \left (x \right )\) gives

\begin{align*} p \left (x \right )&=-{\frac {58}{243}}\\ p \left (x \right )&=-\frac {6^{{1}/{3}} \left (4 \,6^{{1}/{3}}+\sqrt {15}\, 6^{{1}/{3}}\right )^{5}}{162}+\frac {55 \,6^{{2}/{3}} \left (4 \,6^{{1}/{3}}+\sqrt {15}\, 6^{{1}/{3}}\right )}{27}+\frac {56 \,6^{{1}/{3}} \left (4 \,6^{{1}/{3}}+\sqrt {15}\, 6^{{1}/{3}}\right )^{2}}{27}+2 \left (4 \,6^{{1}/{3}}+\sqrt {15}\, 6^{{1}/{3}}\right )^{3} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \int _{}^{p \left (x \right )}\frac {\left (6^{{1}/{3}}+\left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}}\right ) \left (6^{{1}/{3}}-\left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}}\right )}{\left (9 \,6^{{1}/{3}} \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}} \tau -2 \,6^{{2}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}}d \tau &= \sqrt {3}\, \ln \left (\frac {1}{x^{{1}/{54}}}\right )+c_1 \\ p \left (x \right ) &= -{\frac {58}{243}} \\ p \left (x \right ) &= -\frac {6^{{1}/{3}} \left (4 \,6^{{1}/{3}}+\sqrt {15}\, 6^{{1}/{3}}\right )^{5}}{162}+\frac {55 \,6^{{2}/{3}} \left (4 \,6^{{1}/{3}}+\sqrt {15}\, 6^{{1}/{3}}\right )}{27}+\frac {56 \,6^{{1}/{3}} \left (4 \,6^{{1}/{3}}+\sqrt {15}\, 6^{{1}/{3}}\right )^{2}}{27}+2 \left (4 \,6^{{1}/{3}}+\sqrt {15}\, 6^{{1}/{3}}\right )^{3} \\ \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} \text {Expression too large to display} \\ u \left (x \right ) &= \frac {2 x}{9} \\ u \left (x \right ) &= \frac {16 x}{9} \\ \end{align*}

Solving ode 2A

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(u \left (x \right )=x f + g\) to (1A) shows that

\begin{align*} f &= \frac {i \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}} \sqrt {3}-\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}}-4 i \sqrt {3}-4}{18 \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}}\\ g &= 0 \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p -\frac {i \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}} \sqrt {3}-\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}}-4 i \sqrt {3}-4}{18 \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}} = \left (\frac {5832 i x \sqrt {3}\, p}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{4}/{3}} \sqrt {6561 p^{2}+3024 p +348}}+\frac {72 i x \sqrt {3}}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{4}/{3}}}+\frac {18 i x \sqrt {3}}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}}}-\frac {18 x}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}}}-\frac {1458 x p}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}} \sqrt {6561 p^{2}+3024 p +348}}-\frac {336 x}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}} \sqrt {6561 p^{2}+3024 p +348}}+\frac {1458 i x \sqrt {3}\, p}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}} \sqrt {6561 p^{2}+3024 p +348}}+\frac {1344 i x \sqrt {3}}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{4}/{3}} \sqrt {6561 p^{2}+3024 p +348}}+\frac {336 i x \sqrt {3}}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}} \sqrt {6561 p^{2}+3024 p +348}}+\frac {72 x}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{4}/{3}}}+\frac {5832 x p}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{4}/{3}} \sqrt {6561 p^{2}+3024 p +348}}+\frac {1344 x}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{4}/{3}} \sqrt {6561 p^{2}+3024 p +348}}\right ) p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p -\frac {i \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}} \sqrt {3}-\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}}-4 i \sqrt {3}-4}{18 \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}} = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=-{\frac {2}{9}}\\ p_{2} &=\frac {i \sqrt {3}\, \left (\frac {7}{2}+\frac {7 i \sqrt {3}}{2}-\frac {3 \sqrt {10 i \sqrt {3}-10}}{2}\right )^{2}-4 i \sqrt {3}-\left (\frac {7}{2}+\frac {7 i \sqrt {3}}{2}-\frac {3 \sqrt {10 i \sqrt {3}-10}}{2}\right )^{2}-4}{63+63 i \sqrt {3}-27 \sqrt {10 i \sqrt {3}-10}} \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} u \left (x \right ) = -\frac {2 x}{9}\\ u \left (x \right ) = -\frac {14 x}{9} \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )-\frac {i \left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}} \sqrt {3}-\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}}-4 i \sqrt {3}-4}{18 \left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{1}/{3}}}}{\frac {5832 i x \sqrt {3}\, p \left (x \right )}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{4}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}+\frac {72 i x \sqrt {3}}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{4}/{3}}}+\frac {18 i x \sqrt {3}}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}}}-\frac {18 x}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}}}-\frac {1458 x p \left (x \right )}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}-\frac {336 x}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}+\frac {1458 i x \sqrt {3}\, p \left (x \right )}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}+\frac {1344 i x \sqrt {3}}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{4}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}+\frac {336 i x \sqrt {3}}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}+\frac {72 x}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{4}/{3}}}+\frac {5832 x p \left (x \right )}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{4}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}+\frac {1344 x}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{4}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

The ode

\begin{equation} p^{\prime }\left (x \right ) = \frac {\sqrt {3}\, \left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i p \left (x \right ) \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}-2 \,3^{{2}/{3}} 2^{{2}/{3}}+9 p \left (x \right ) \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}+4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}}{27 \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}}\right ) x} \end{equation}

is separable as it can be written as

\begin{align*} p^{\prime }\left (x \right )&= \frac {\sqrt {3}\, \left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i p \left (x \right ) \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}-2 \,3^{{2}/{3}} 2^{{2}/{3}}+9 p \left (x \right ) \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}+4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}}{27 \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}}\right ) x}\\ &= f(x) g(p) \end{align*}

Where

\begin{align*} f(x) &= \frac {\sqrt {3}}{27 x}\\ g(p) &= \frac {\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i p \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}-2 \,3^{{2}/{3}} 2^{{2}/{3}}+9 p \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}+4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}}{\left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right )} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(x) \,dx} \\ \int { \frac {\left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right )}{\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i p \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}-2 \,3^{{2}/{3}} 2^{{2}/{3}}+9 p \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}+4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}}\,dp} &= \int { \frac {\sqrt {3}}{27 x} \,dx} \\ \end{align*}

\[ \int _{}^{p \left (x \right )}\frac {\left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}}\right )}{\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \tau \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}-2 \,3^{{2}/{3}} 2^{{2}/{3}}+9 \tau \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}+4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}}d \tau = \sqrt {3}\, \ln \left (x^{{1}/{27}}\right )+c_2 \]

We now need to find the singular solutions, these are found by finding for what values \(g(p)\) is zero, since we had to divide by this above. Solving \(g(p)=0\) or

\[ \frac {\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i p \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}-2 \,3^{{2}/{3}} 2^{{2}/{3}}+9 p \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}+4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}}{\left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right )}=0 \]

for \(p \left (x \right )\) gives

\begin{align*} p \left (x \right )&=-{\frac {2}{9}}\\ p \left (x \right )&=\frac {3^{{1}/{3}} 2^{{1}/{3}} \left (i 3^{{5}/{6}} 2^{{1}/{3}} \left (\frac {7 i 3^{{5}/{6}} 2^{{1}/{3}}}{4}+\frac {7 \,3^{{1}/{3}} 2^{{1}/{3}}}{4}-\frac {3 \sqrt {-10 \,3^{{2}/{3}} 2^{{2}/{3}}+30 i 3^{{1}/{6}} 2^{{2}/{3}}}}{4}\right )^{2}-3^{{1}/{3}} 2^{{1}/{3}} \left (\frac {7 i 3^{{5}/{6}} 2^{{1}/{3}}}{4}+\frac {7 \,3^{{1}/{3}} 2^{{1}/{3}}}{4}-\frac {3 \sqrt {-10 \,3^{{2}/{3}} 2^{{2}/{3}}+30 i 3^{{1}/{6}} 2^{{2}/{3}}}}{4}\right )^{2}-6 i \sqrt {3}-6\right )}{\frac {189 i 3^{{5}/{6}} 2^{{1}/{3}}}{2}+\frac {189 \,3^{{1}/{3}} 2^{{1}/{3}}}{2}-\frac {81 \sqrt {-10 \,3^{{2}/{3}} 2^{{2}/{3}}+30 i 3^{{1}/{6}} 2^{{2}/{3}}}}{2}} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \int _{}^{p \left (x \right )}\frac {\left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}-2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}}\right )}{\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \tau \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}-2 \,3^{{2}/{3}} 2^{{2}/{3}}+9 \tau \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}+4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}}d \tau &= \sqrt {3}\, \ln \left (x^{{1}/{27}}\right )+c_2 \\ p \left (x \right ) &= -{\frac {2}{9}} \\ p \left (x \right ) &= \frac {3^{{1}/{3}} 2^{{1}/{3}} \left (i 3^{{5}/{6}} 2^{{1}/{3}} \left (\frac {7 i 3^{{5}/{6}} 2^{{1}/{3}}}{4}+\frac {7 \,3^{{1}/{3}} 2^{{1}/{3}}}{4}-\frac {3 \sqrt {-10 \,3^{{2}/{3}} 2^{{2}/{3}}+30 i 3^{{1}/{6}} 2^{{2}/{3}}}}{4}\right )^{2}-3^{{1}/{3}} 2^{{1}/{3}} \left (\frac {7 i 3^{{5}/{6}} 2^{{1}/{3}}}{4}+\frac {7 \,3^{{1}/{3}} 2^{{1}/{3}}}{4}-\frac {3 \sqrt {-10 \,3^{{2}/{3}} 2^{{2}/{3}}+30 i 3^{{1}/{6}} 2^{{2}/{3}}}}{4}\right )^{2}-6 i \sqrt {3}-6\right )}{\frac {189 i 3^{{5}/{6}} 2^{{1}/{3}}}{2}+\frac {189 \,3^{{1}/{3}} 2^{{1}/{3}}}{2}-\frac {81 \sqrt {-10 \,3^{{2}/{3}} 2^{{2}/{3}}+30 i 3^{{1}/{6}} 2^{{2}/{3}}}}{2}} \\ \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} \text {Expression too large to display} \\ u \left (x \right ) &= -\frac {2 x}{9} \\ u \left (x \right ) &= -\frac {14 x}{9} \\ \end{align*}

Solving ode 3A

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(u \left (x \right )=x f + g\) to (1A) shows that

\begin{align*} f &= -\frac {\left (1+i \sqrt {3}\right ) \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}}-4 i \sqrt {3}+4}{18 \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}}\\ g &= 0 \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p +\frac {\left (1+i \sqrt {3}\right ) \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}}-4 i \sqrt {3}+4}{18 \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}} = \left (-\frac {18 x}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}}}-\frac {1458 x p}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}} \sqrt {6561 p^{2}+3024 p +348}}-\frac {336 x}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}} \sqrt {6561 p^{2}+3024 p +348}}-\frac {18 i x \sqrt {3}}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}}}-\frac {336 i x \sqrt {3}}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}} \sqrt {6561 p^{2}+3024 p +348}}-\frac {1458 i x \sqrt {3}\, p}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}} \sqrt {6561 p^{2}+3024 p +348}}-\frac {72 i x \sqrt {3}}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{4}/{3}}}-\frac {1344 i x \sqrt {3}}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{4}/{3}} \sqrt {6561 p^{2}+3024 p +348}}-\frac {5832 i x \sqrt {3}\, p}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{4}/{3}} \sqrt {6561 p^{2}+3024 p +348}}+\frac {72 x}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{4}/{3}}}+\frac {5832 x p}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{4}/{3}} \sqrt {6561 p^{2}+3024 p +348}}+\frac {1344 x}{\left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{4}/{3}} \sqrt {6561 p^{2}+3024 p +348}}\right ) p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} p +\frac {\left (1+i \sqrt {3}\right ) \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{2}/{3}}-4 i \sqrt {3}+4}{18 \left (224+972 p +12 \sqrt {6561 p^{2}+3024 p +348}\right )^{{1}/{3}}} = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=-{\frac {2}{9}} \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} u \left (x \right ) = -\frac {2 x}{9} \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )+\frac {\left (1+i \sqrt {3}\right ) \left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}}-4 i \sqrt {3}+4}{18 \left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{1}/{3}}}}{-\frac {18 x}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}}}-\frac {1458 x p \left (x \right )}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}-\frac {336 x}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}-\frac {18 i x \sqrt {3}}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}}}-\frac {336 i x \sqrt {3}}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}-\frac {1458 i x \sqrt {3}\, p \left (x \right )}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{2}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}-\frac {72 i x \sqrt {3}}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{4}/{3}}}-\frac {1344 i x \sqrt {3}}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{4}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}-\frac {5832 i x \sqrt {3}\, p \left (x \right )}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{4}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}+\frac {72 x}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{4}/{3}}}+\frac {5832 x p \left (x \right )}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{4}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}+\frac {1344 x}{\left (224+972 p \left (x \right )+12 \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}\right )^{{4}/{3}} \sqrt {6561 p \left (x \right )^{2}+3024 p \left (x \right )+348}}} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

The ode

\begin{equation} p^{\prime }\left (x \right ) = \frac {\sqrt {3}\, \left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i p \left (x \right ) \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}+2 \,3^{{2}/{3}} 2^{{2}/{3}}-9 p \left (x \right ) \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}-4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}}{27 \left (-i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}-3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}}\right ) x} \end{equation}

is separable as it can be written as

\begin{align*} p^{\prime }\left (x \right )&= \frac {\sqrt {3}\, \left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i p \left (x \right ) \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}+2 \,3^{{2}/{3}} 2^{{2}/{3}}-9 p \left (x \right ) \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}-4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}}{27 \left (-i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}-3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p \left (x \right )+56 \sqrt {3}+9 \sqrt {\left (243 p \left (x \right )+58\right ) \left (9 p \left (x \right )+2\right )}\right )\right )^{{1}/{3}}\right ) x}\\ &= f(x) g(p) \end{align*}

Where

\begin{align*} f(x) &= \frac {\sqrt {3}}{27 x}\\ g(p) &= \frac {\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i p \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}+2 \,3^{{2}/{3}} 2^{{2}/{3}}-9 p \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}-4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}}{\left (-i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}-3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right )} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(x) \,dx} \\ \int { \frac {\left (-i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}-3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right )}{\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i p \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}+2 \,3^{{2}/{3}} 2^{{2}/{3}}-9 p \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}-4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}}\,dp} &= \int { \frac {\sqrt {3}}{27 x} \,dx} \\ \end{align*}

\[ \int _{}^{p \left (x \right )}\frac {\left (-i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}-3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}}\right )}{\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \tau \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}+2 \,3^{{2}/{3}} 2^{{2}/{3}}-9 \tau \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}-4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}}d \tau = \sqrt {3}\, \ln \left (x^{{1}/{27}}\right )+c_3 \]

We now need to find the singular solutions, these are found by finding for what values \(g(p)\) is zero, since we had to divide by this above. Solving \(g(p)=0\) or

\[ \frac {\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i p \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}+2 \,3^{{2}/{3}} 2^{{2}/{3}}-9 p \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}-4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}}{\left (-i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}-3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, p +56 \sqrt {3}+9 \sqrt {\left (243 p +58\right ) \left (9 p +2\right )}\right )\right )^{{1}/{3}}\right )}=0 \]

for \(p \left (x \right )\) gives

\begin{align*} p \left (x \right )&=-{\frac {58}{243}}\\ p \left (x \right )&=-{\frac {2}{9}} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \int _{}^{p \left (x \right )}\frac {\left (-i 3^{{5}/{6}} 2^{{1}/{3}}+3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}}\right ) \left (i 3^{{5}/{6}} 2^{{1}/{3}}-3^{{1}/{3}} 2^{{1}/{3}}+2 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}}\right )}{\left (6 i 3^{{1}/{6}} 2^{{2}/{3}}+9 i \tau \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}} 3^{{5}/{6}} 2^{{1}/{3}}+2 \,3^{{2}/{3}} 2^{{2}/{3}}-9 \tau \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{1}/{3}} 3^{{1}/{3}} 2^{{1}/{3}}-4 \left (\sqrt {3}\, \left (243 \sqrt {3}\, \tau +56 \sqrt {3}+9 \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}\right )\right )^{{2}/{3}}\right ) \sqrt {\left (243 \tau +58\right ) \left (9 \tau +2\right )}}d \tau &= \sqrt {3}\, \ln \left (x^{{1}/{27}}\right )+c_3 \\ p \left (x \right ) &= -{\frac {58}{243}} \\ p \left (x \right ) &= -{\frac {2}{9}} \\ \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} \text {Expression too large to display} \\ u \left (x \right ) &= \frac {2 x}{9} \\ u \left (x \right ) &= -\frac {2 x}{9} \\ \end{align*}

The solution \(\text {Expression too large to display}\) simplifies to

\begin{align*} \text {Expression too large to display} \\ \end{align*}

The solution \(\text {Expression too large to display}\) simplifies to

\begin{align*} \text {Expression too large to display} \\ \end{align*}

The solution \(\text {Expression too large to display}\) simplifies to

\begin{align*} \text {Expression too large to display} \\ \end{align*}

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

The above solution was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

\begin{align*} u \left (x \right ) &= -\frac {14 x}{9} \\ u \left (x \right ) &= -\frac {2 x}{9} \\ u \left (x \right ) &= \frac {16 x}{9} \\ \end{align*}

2.1.16.16 Solved using first_order_ode_homog_type_D2

1.938 (sec)

Entering first order ode homog type D2 solver

\begin{align*} u^{\prime }\left (x \right )&=\frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )}{18 x}-\frac {56}{243} \\ \end{align*}

Applying change of variables \(u \left (x \right ) = u \left (x \right ) x\), then the ode becomes

\begin{align*} u^{\prime }\left (x \right ) x +u \left (x \right ) = \frac {3 u \left (x \right )^{3}}{8}-\frac {u \left (x \right )}{18}-\frac {56}{243} \end{align*}

Which is now solved The ode

\begin{equation} u^{\prime }\left (x \right ) = \frac {\left (9 u \left (x \right )+2\right ) \left (9 u \left (x \right )-16\right ) \left (9 u \left (x \right )+14\right )}{1944 x} \end{equation}

is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= \frac {\left (9 u \left (x \right )+2\right ) \left (9 u \left (x \right )-16\right ) \left (9 u \left (x \right )+14\right )}{1944 x}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{1944 x}\\ g(u) &= \left (9 u +2\right ) \left (9 u -16\right ) \left (9 u +14\right ) \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\ \int { \frac {1}{\left (9 u +2\right ) \left (9 u -16\right ) \left (9 u +14\right )}\,du} &= \int { \frac {1}{1944 x} \,dx} \\ \end{align*}

\[ \frac {\ln \left (9 u \left (x \right )-16\right )}{4860}+\frac {\ln \left (9 u \left (x \right )+14\right )}{3240}-\frac {\ln \left (9 u \left (x \right )+2\right )}{1944}=\ln \left (x^{\frac {1}{1944}}\right )+c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ \left (9 u +2\right ) \left (9 u -16\right ) \left (9 u +14\right )=0 \]

for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=-{\frac {14}{9}}\\ u \left (x \right )&=-{\frac {2}{9}} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \frac {\ln \left (9 u \left (x \right )-16\right )}{4860}+\frac {\ln \left (9 u \left (x \right )+14\right )}{3240}-\frac {\ln \left (9 u \left (x \right )+2\right )}{1944} &= \ln \left (x^{\frac {1}{1944}}\right )+c_1 \\ u \left (x \right ) &= -{\frac {14}{9}} \\ u \left (x \right ) &= -{\frac {2}{9}} \\ \end{align*}

Converting \(\frac {\ln \left (9 u \left (x \right )-16\right )}{4860}+\frac {\ln \left (9 u \left (x \right )+14\right )}{3240}-\frac {\ln \left (9 u \left (x \right )+2\right )}{1944} = \ln \left (x^{\frac {1}{1944}}\right )+c_1\) back to \(u \left (x \right )\) gives

\begin{align*} \frac {\ln \left (\frac {9 u \left (x \right )}{x}-16\right )}{4860}+\frac {\ln \left (\frac {9 u \left (x \right )}{x}+14\right )}{3240}-\frac {\ln \left (\frac {9 u \left (x \right )}{x}+2\right )}{1944} = \ln \left (x^{\frac {1}{1944}}\right )+c_1 \end{align*}

Converting \(u \left (x \right ) = -{\frac {14}{9}}\) back to \(u \left (x \right )\) gives

\begin{align*} u \left (x \right ) = -\frac {14 x}{9} \end{align*}

Converting \(u \left (x \right ) = -{\frac {2}{9}}\) back to \(u \left (x \right )\) gives

\begin{align*} u \left (x \right ) = -\frac {2 x}{9} \end{align*}

Simplifying the above gives

\begin{align*} \frac {\ln \left (\frac {9 u \left (x \right )-16 x}{x}\right )}{4860}+\frac {\ln \left (\frac {9 u \left (x \right )+14 x}{x}\right )}{3240}-\frac {\ln \left (\frac {9 u \left (x \right )+2 x}{x}\right )}{1944} &= \frac {\ln \left (x \right )}{1944}+c_1 \\ u \left (x \right ) &= -\frac {14 x}{9} \\ u \left (x \right ) &= -\frac {2 x}{9} \\ \end{align*}

Summary of solutions found

\begin{align*} \frac {\ln \left (\frac {9 u \left (x \right )-16 x}{x}\right )}{4860}+\frac {\ln \left (\frac {9 u \left (x \right )+14 x}{x}\right )}{3240}-\frac {\ln \left (\frac {9 u \left (x \right )+2 x}{x}\right )}{1944} &= \frac {\ln \left (x \right )}{1944}+c_1 \\ u \left (x \right ) &= -\frac {14 x}{9} \\ u \left (x \right ) &= -\frac {2 x}{9} \\ \end{align*}

2.1.16.17 Solved using first_order_ode_homog_type_G

1.625 (sec)

Entering first order ode homog type G solver

\begin{align*} u^{\prime }\left (x \right )&=\frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )}{18 x}-\frac {56}{243} \\ \end{align*}

Multiplying the right side of the ode, which is \(\frac {3 u^{3}}{8 x^{3}}-\frac {u}{18 x}-\frac {56}{243}\) by \(\frac {x}{u}\) gives

\begin{align*} u^{\prime }\left (x \right ) &= \left (\frac {x}{u}\right ) \frac {3 u^{3}}{8 x^{3}}-\frac {u}{18 x}-\frac {56}{243}\\ &= \frac {729 u^{3}-108 u \,x^{2}-448 x^{3}}{1944 x^{2} u}\\ &= F(x,u) \end{align*}

Since \(F \left (x , u\right )\) has \(u\), then let

\begin{align*} f_x&= x \left (\frac {\partial }{\partial x}F \left (x , u\right )\right )\\ &= -\frac {729 u^{3}+224 x^{3}}{972 x^{2} u}\\ f_u&= u \left (\frac {\partial }{\partial u}F \left (x , u\right )\right )\\ &= \frac {729 u^{3}+224 x^{3}}{972 x^{2} u}\\ \alpha &= \frac {f_x}{f_u} \\ &=-1 \end{align*}

Since \(\alpha \) is independent of \(x,u\) then this is Homogeneous type G.

Let

\begin{align*} u&=\frac {z}{x^ \alpha }\\ &=\frac {z}{\frac {1}{x}} \end{align*}

Substituting the above back into \(F(x,u)\) gives

\begin{align*} F \left (z \right ) &=\frac {729 z^{3}-108 z -448}{1944 z} \end{align*}

We see that \(F \left (z \right )\) does not depend on \(x\) nor on \(u\). If this was not the case, then this method will not work.

Therefore, the implicit solution is given by

\begin{align*} \ln \left (x \right )- c_1 - \int ^{u x^\alpha } \frac {1}{z \left (\alpha + F(z)\right ) } \,dz & = 0 \end{align*}

Which gives

\[ \ln \left (x \right )-c_1 +\int _{}^{\frac {u \left (x \right )}{x}}\frac {1}{z \left (1-\frac {729 z^{3}-108 z -448}{1944 z}\right )}d z = 0 \]

The value of the above is

\[ \ln \left (x \right )-c_1 -\frac {2 \ln \left (\frac {9 u \left (x \right )-16 x}{x}\right )}{5}-\frac {3 \ln \left (\frac {9 u \left (x \right )+14 x}{x}\right )}{5}+\ln \left (\frac {9 u \left (x \right )+2 x}{x}\right ) = 0 \]

Summary of solutions found

\begin{align*} \ln \left (x \right )-c_1 -\frac {2 \ln \left (\frac {9 u \left (x \right )-16 x}{x}\right )}{5}-\frac {3 \ln \left (\frac {9 u \left (x \right )+14 x}{x}\right )}{5}+\ln \left (\frac {9 u \left (x \right )+2 x}{x}\right ) &= 0 \\ \end{align*}

2.1.16.18 Solved using first_order_ode_homog_type_maple_C

12.882 (sec)

Entering first order ode homog type maple C solver

\begin{align*} u^{\prime }\left (x \right )&=\frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )}{18 x}-\frac {56}{243} \\ \end{align*}

Let \(Y = u -y_{0}\) and \(X = x -x_{0}\) then the above is transformed to new ode in \(Y(X)\)

\[ \frac {d}{d X}Y \left (X \right ) = \frac {729 \left (Y \left (X \right )+y_{0} \right )^{3}-108 \left (Y \left (X \right )+y_{0} \right ) \left (X +x_{0} \right )^{2}-448 \left (X +x_{0} \right )^{3}}{1944 \left (X +x_{0} \right )^{3}} \]

Solving for possible values of \(x_{0}\) and \(y_{0}\) which makes the above ode a homogeneous ode results in

\begin{align*} x_{0}&=0\\ y_{0}&=0 \end{align*}

Using these values now it is possible to easily solve for \(Y \left (X \right )\). The above ode now becomes

\begin{align*} \frac {d}{d X}Y \left (X \right ) = \frac {-448 X^{3}-108 Y \left (X \right ) X^{2}+729 Y \left (X \right )^{3}}{1944 X^{3}} \end{align*}

In canonical form, the ODE is

\begin{align*} Y' &= F(X,Y)\\ &= \frac {-448 X^{3}-108 Y \,X^{2}+729 Y^{3}}{1944 X^{3}}\tag {1} \end{align*}

An ode of the form \(Y' = \frac {M(X,Y)}{N(X,Y)}\) is called homogeneous if the functions \(M(X,Y)\) and \(N(X,Y)\) are both homogeneous functions and of the same order. Recall that a function \(f(X,Y)\) is homogeneous of order \(n\) if

\[ f(t^n X, t^n Y)= t^n f(X,Y) \]

In this case, it can be seen that both \(M=-448 X^{3}-108 Y \,X^{2}+729 Y^{3}\) and \(N=1944 X^{3}\) are both homogeneous and of the same order \(n=3\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {Y}{X}\), or \(Y=uX\). Hence

\[ \frac { \mathop {\mathrm {d}Y}}{\mathop {\mathrm {d}X}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}}X + u \]

Applying the transformation \(Y=uX\) to the above ODE in (1) gives

\begin{align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}}X + u &= -\frac {56}{243}-\frac {1}{18} u +\frac {3}{8} u^{3}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {-\frac {56}{243}-\frac {19 u \left (X \right )}{18}+\frac {3 u \left (X \right )^{3}}{8}}{X} \end{align*}

Or

\[ \frac {d}{d X}u \left (X \right )-\frac {-\frac {56}{243}-\frac {19 u \left (X \right )}{18}+\frac {3 u \left (X \right )^{3}}{8}}{X} = 0 \]

Or

\[ -729 u \left (X \right )^{3}+1944 \left (\frac {d}{d X}u \left (X \right )\right ) X +2052 u \left (X \right )+448 = 0 \]

Which is now solved as separable in \(u \left (X \right )\).

The ode

\begin{equation} \frac {d}{d X}u \left (X \right ) = \frac {\left (9 u \left (X \right )+2\right ) \left (9 u \left (X \right )-16\right ) \left (9 u \left (X \right )+14\right )}{1944 X} \end{equation}

is separable as it can be written as

\begin{align*} \frac {d}{d X}u \left (X \right )&= \frac {\left (9 u \left (X \right )+2\right ) \left (9 u \left (X \right )-16\right ) \left (9 u \left (X \right )+14\right )}{1944 X}\\ &= f(X) g(u) \end{align*}

Where

\begin{align*} f(X) &= \frac {1}{1944 X}\\ g(u) &= \left (9 u +2\right ) \left (9 u -16\right ) \left (9 u +14\right ) \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(X) \,dX} \\ \int { \frac {1}{\left (9 u +2\right ) \left (9 u -16\right ) \left (9 u +14\right )}\,du} &= \int { \frac {1}{1944 X} \,dX} \\ \end{align*}

\[ \frac {\ln \left (9 u \left (X \right )-16\right )}{4860}+\frac {\ln \left (9 u \left (X \right )+14\right )}{3240}-\frac {\ln \left (9 u \left (X \right )+2\right )}{1944}=\ln \left (X^{\frac {1}{1944}}\right )+c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ \left (9 u +2\right ) \left (9 u -16\right ) \left (9 u +14\right )=0 \]

for \(u \left (X \right )\) gives

\begin{align*} u \left (X \right )&=-{\frac {14}{9}}\\ u \left (X \right )&=-{\frac {2}{9}} \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \frac {\ln \left (9 u \left (X \right )-16\right )}{4860}+\frac {\ln \left (9 u \left (X \right )+14\right )}{3240}-\frac {\ln \left (9 u \left (X \right )+2\right )}{1944} &= \ln \left (X^{\frac {1}{1944}}\right )+c_1 \\ u \left (X \right ) &= -{\frac {14}{9}} \\ u \left (X \right ) &= -{\frac {2}{9}} \\ \end{align*}

Converting \(\frac {\ln \left (9 u \left (X \right )-16\right )}{4860}+\frac {\ln \left (9 u \left (X \right )+14\right )}{3240}-\frac {\ln \left (9 u \left (X \right )+2\right )}{1944} = \ln \left (X^{\frac {1}{1944}}\right )+c_1\) back to \(Y \left (X \right )\) gives

\begin{align*} \frac {2 \ln \left (\frac {9 Y \left (X \right )-16 X}{X}\right )}{5}+\frac {3 \ln \left (\frac {9 Y \left (X \right )+14 X}{X}\right )}{5}-\ln \left (\frac {9 Y \left (X \right )+2 X}{X}\right ) = \ln \left (X \right )+1944 c_1 \end{align*}

Converting \(u \left (X \right ) = -{\frac {14}{9}}\) back to \(Y \left (X \right )\) gives

\begin{align*} Y \left (X \right ) = -\frac {14 X}{9} \end{align*}

Converting \(u \left (X \right ) = -{\frac {2}{9}}\) back to \(Y \left (X \right )\) gives

\begin{align*} Y \left (X \right ) = -\frac {2 X}{9} \end{align*}

Using the solution for \(Y(X)\)

\begin{align*} \frac {2 \ln \left (\frac {9 Y \left (X \right )-16 X}{X}\right )}{5}+\frac {3 \ln \left (\frac {9 Y \left (X \right )+14 X}{X}\right )}{5}-\ln \left (\frac {9 Y \left (X \right )+2 X}{X}\right ) = \ln \left (X \right )+1944 c_1\tag {A} \end{align*}

And replacing back terms in the above solution using

\begin{align*} Y &= u +y_{0}\\ X &= x +x_{0} \end{align*}

Or

\begin{align*} Y &= u\\ X &= x \end{align*}

Then the solution in \(u \left (x \right )\) becomes using EQ (A)

\begin{align*} \frac {2 \ln \left (\frac {9 u \left (x \right )-16 x}{x}\right )}{5}+\frac {3 \ln \left (\frac {9 u \left (x \right )+14 x}{x}\right )}{5}-\ln \left (\frac {9 u \left (x \right )+2 x}{x}\right ) = \ln \left (x \right )+1944 c_1 \end{align*}

Using the solution for \(Y(X)\)

\begin{align*} Y \left (X \right ) = -\frac {14 X}{9}\tag {A} \end{align*}

And replacing back terms in the above solution using

\begin{align*} Y &= u +y_{0}\\ X &= x +x_{0} \end{align*}

Or

\begin{align*} Y &= u\\ X &= x \end{align*}

Then the solution in \(u \left (x \right )\) becomes using EQ (A)

\begin{align*} u \left (x \right ) = -\frac {14 x}{9} \end{align*}

Using the solution for \(Y(X)\)

\begin{align*} Y \left (X \right ) = -\frac {2 X}{9}\tag {A} \end{align*}

And replacing back terms in the above solution using

\begin{align*} Y &= u +y_{0}\\ X &= x +x_{0} \end{align*}

Or

\begin{align*} Y &= u\\ X &= x \end{align*}

Then the solution in \(u \left (x \right )\) becomes using EQ (A)

\begin{align*} u \left (x \right ) = -\frac {2 x}{9} \end{align*}

Summary of solutions found

\begin{align*} \frac {2 \ln \left (\frac {9 u \left (x \right )-16 x}{x}\right )}{5}+\frac {3 \ln \left (\frac {9 u \left (x \right )+14 x}{x}\right )}{5}-\ln \left (\frac {9 u \left (x \right )+2 x}{x}\right ) &= \ln \left (x \right )+1944 c_1 \\ u \left (x \right ) &= -\frac {14 x}{9} \\ u \left (x \right ) &= -\frac {2 x}{9} \\ \end{align*}

2.1.16.19 Solved using first_order_ode_chini

0.129 (sec)

Entering first order ode chini solver

\begin{align*} u^{\prime }\left (x \right )&=\frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )}{18 x}-\frac {56}{243} \\ \end{align*}

The solution to this Chini ode is (more steps will be added showing how soon).

\[ \int _{}^{-\frac {513 u \left (x \right )}{112 x}}\frac {1}{\frac {3136}{185193} u^{3}-u +1}d u +\frac {152 x^{3} \ln \left (x \right ) 7^{{2}/{3}}}{729 \left (-\frac {448 x^{3}}{729}\right )^{{2}/{3}} \left (-x^{3}\right )^{{1}/{3}}}+c_1 = 0 \]

Summary of solutions found

\begin{align*} \int _{}^{-\frac {513 u \left (x \right )}{112 x}}\frac {1}{\frac {3136}{185193} u^{3}-u +1}d u +\frac {152 x^{3} \ln \left (x \right ) 7^{{2}/{3}}}{729 \left (-\frac {448 x^{3}}{729}\right )^{{2}/{3}} \left (-x^{3}\right )^{{1}/{3}}}+c_1 &= 0 \\ \end{align*}

2.1.16.20 Solved using first_order_ode_LIE

6.689 (sec)

Entering first order ode LIE solver

\begin{align*} u^{\prime }\left (x \right )&=\frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )}{18 x}-\frac {56}{243} \\ \end{align*}

Writing the ode as

\begin{align*} u^{\prime }\left (x \right )&=\frac {729 u^{3}-108 u \,x^{2}-448 x^{3}}{1944 x^{3}}\\ u^{\prime }\left (x \right )&= \omega \left ( x,u \left (x \right )\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

\begin{align*} \eta _{x}+\omega \left ( \eta _{u \left (x \right )}-\xi _{x}\right ) -\omega ^{2}\xi _{u \left (x \right )}-\omega _{x}\xi -\omega _{u \left (x \right )}\eta =0\tag {A} \end{align*}

To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives

\begin{align*} \tag{1E} \xi &= u a_{3}+x a_{2}+a_{1} \\ \tag{2E} \eta &= u b_{3}+x b_{2}+b_{1} \\ \end{align*}

Where the unknown coefficients are

\[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \]

Substituting equations (1E,2E) and \(\omega \) into (A) gives

\begin{equation} \tag{5E} b_{2}+\frac {\left (729 u^{3}-108 u \,x^{2}-448 x^{3}\right ) \left (b_{3}-a_{2}\right )}{1944 x^{3}}-\frac {\left (729 u^{3}-108 u \,x^{2}-448 x^{3}\right )^{2} a_{3}}{3779136 x^{6}}-\left (\frac {-216 u x -1344 x^{2}}{1944 x^{3}}-\frac {729 u^{3}-108 u \,x^{2}-448 x^{3}}{648 x^{4}}\right ) \left (u a_{3}+x a_{2}+a_{1}\right )-\frac {\left (2187 u^{2}-108 x^{2}\right ) \left (u b_{3}+x b_{2}+b_{1}\right )}{1944 x^{3}} = 0 \end{equation}

Putting the above in normal form gives

\[ -\frac {531441 u^{6} a_{3}-4408992 u^{4} x^{2} a_{3}-2834352 u^{3} x^{3} a_{2}-653184 u^{3} x^{3} a_{3}+2834352 u^{3} x^{3} b_{3}+221616 u^{2} x^{4} a_{3}+4251528 u^{2} x^{4} b_{2}+96768 u \,x^{5} a_{3}-870912 x^{6} a_{2}+200704 x^{6} a_{3}-3989088 b_{2} x^{6}+870912 x^{6} b_{3}-4251528 u^{3} x^{2} a_{1}+4251528 u^{2} x^{3} b_{1}+209952 u \,x^{4} a_{1}-209952 x^{5} b_{1}}{3779136 x^{6}} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} -531441 u^{6} a_{3}+4408992 u^{4} x^{2} a_{3}+2834352 u^{3} x^{3} a_{2}+653184 u^{3} x^{3} a_{3}-2834352 u^{3} x^{3} b_{3}-221616 u^{2} x^{4} a_{3}-4251528 u^{2} x^{4} b_{2}-96768 u \,x^{5} a_{3}+870912 x^{6} a_{2}-200704 x^{6} a_{3}+3989088 b_{2} x^{6}-870912 x^{6} b_{3}+4251528 u^{3} x^{2} a_{1}-4251528 u^{2} x^{3} b_{1}-209952 u \,x^{4} a_{1}+209952 x^{5} b_{1} = 0 \end{equation}

Looking at the above PDE shows the following are all the terms with \(\{u, x\}\) in them.

\[ \{u, x\} \]

The following substitution is now made to be able to collect on all terms with \(\{u, x\}\) in them

\[ \{u = v_{1}, x = v_{2}\} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} 2834352 a_{2} v_{1}^{3} v_{2}^{3}+870912 a_{2} v_{2}^{6}-531441 a_{3} v_{1}^{6}+4408992 a_{3} v_{1}^{4} v_{2}^{2}+653184 a_{3} v_{1}^{3} v_{2}^{3}-221616 a_{3} v_{1}^{2} v_{2}^{4}-96768 a_{3} v_{1} v_{2}^{5}-200704 a_{3} v_{2}^{6}-4251528 b_{2} v_{1}^{2} v_{2}^{4}+3989088 b_{2} v_{2}^{6}-2834352 b_{3} v_{1}^{3} v_{2}^{3}-870912 b_{3} v_{2}^{6}+4251528 a_{1} v_{1}^{3} v_{2}^{2}-209952 a_{1} v_{1} v_{2}^{4}-4251528 b_{1} v_{1}^{2} v_{2}^{3}+209952 b_{1} v_{2}^{5} = 0 \end{equation}

Collecting the above on the terms \(v_i\) introduced, and these are

\[ \{v_{1}, v_{2}\} \]

Equation (7E) now becomes

\begin{equation} \tag{8E} -531441 a_{3} v_{1}^{6}+4408992 a_{3} v_{1}^{4} v_{2}^{2}+\left (2834352 a_{2}+653184 a_{3}-2834352 b_{3}\right ) v_{1}^{3} v_{2}^{3}+4251528 a_{1} v_{1}^{3} v_{2}^{2}+\left (-221616 a_{3}-4251528 b_{2}\right ) v_{1}^{2} v_{2}^{4}-4251528 b_{1} v_{1}^{2} v_{2}^{3}-96768 a_{3} v_{1} v_{2}^{5}-209952 a_{1} v_{1} v_{2}^{4}+\left (870912 a_{2}-200704 a_{3}+3989088 b_{2}-870912 b_{3}\right ) v_{2}^{6}+209952 b_{1} v_{2}^{5} = 0 \end{equation}

Setting each coefficients in (8E) to zero gives the following equations to solve

\begin{align*} -209952 a_{1}&=0\\ 4251528 a_{1}&=0\\ -531441 a_{3}&=0\\ -96768 a_{3}&=0\\ 4408992 a_{3}&=0\\ -4251528 b_{1}&=0\\ 209952 b_{1}&=0\\ -221616 a_{3}-4251528 b_{2}&=0\\ 2834352 a_{2}+653184 a_{3}-2834352 b_{3}&=0\\ 870912 a_{2}-200704 a_{3}+3989088 b_{2}-870912 b_{3}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=0\\ a_{2}&=b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \end{align*}

Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives

\begin{align*} \xi &= x \\ \eta &= u \\ \end{align*}

Shifting is now applied to make \(\xi =0\) in order to simplify the rest of the computation

\begin{align*} \eta &= \eta - \omega \left (x,u\right ) \xi \\ &= u - \left (\frac {729 u^{3}-108 u \,x^{2}-448 x^{3}}{1944 x^{3}}\right ) \left (x\right ) \\ &= \frac {-729 u^{3}+2052 u \,x^{2}+448 x^{3}}{1944 x^{2}}\\ \xi &= 0 \end{align*}

The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,u\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.

The characteristic pde which is used to find the canonical coordinates is

\begin{align*} \frac {d x}{\xi } &= \frac {d u}{\eta } = dS \tag {1} \end{align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial u}\right ) S(x,u) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since \(\xi =0\) then in this special case

\begin{align*} R = x \end{align*}

\(S\) is found from

\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{\frac {-729 u^{3}+2052 u \,x^{2}+448 x^{3}}{1944 x^{2}}}} dy \end{align*}

Which results in

\begin{align*} S&= -\frac {3 \ln \left (9 u +14 x \right )}{5}-\frac {2 \ln \left (9 u -16 x \right )}{5}+\ln \left (9 u +2 x \right ) \end{align*}

Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating

\begin{align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,u) S_{u} }{ R_{x} + \omega (x,u) R_{u} }\tag {2} \end{align*}

Where in the above \(R_{x},R_{u},S_{x},S_{u}\) are all partial derivatives and \(\omega (x,u)\) is the right hand side of the original ode given by

\begin{align*} \omega (x,u) &= \frac {729 u^{3}-108 u \,x^{2}-448 x^{3}}{1944 x^{3}} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= 1\\ R_{u} &= 0\\ S_{x} &= \frac {1944 x u}{729 u^{3}-2052 u \,x^{2}-448 x^{3}}\\ S_{u} &= -\frac {1944 x^{2}}{729 u^{3}-2052 u \,x^{2}-448 x^{3}} \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= -\frac {1}{x}\tag {2A} \end{align*}

We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,u\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives

\begin{align*} \frac {dS}{dR} &= -\frac {1}{R} \end{align*}

The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\).

Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {-\frac {1}{R}\, dR}\\ S \left (R \right ) &= -\ln \left (R \right ) + c_2 \end{align*}

\begin{align*} S \left (R \right )&= -\ln \left (R \right )+c_2 \end{align*}

To complete the solution, we just need to transform the above back to \(x,u\) coordinates. This results in

\begin{align*} -\frac {3 \ln \left (9 u \left (x \right )+14 x \right )}{5}-\frac {2 \ln \left (9 u \left (x \right )-16 x \right )}{5}+\ln \left (9 u \left (x \right )+2 x \right ) = -\ln \left (x \right )+c_2 \end{align*}

Summary of solutions found

\begin{align*} -\frac {3 \ln \left (9 u \left (x \right )+14 x \right )}{5}-\frac {2 \ln \left (9 u \left (x \right )-16 x \right )}{5}+\ln \left (9 u \left (x \right )+2 x \right ) &= -\ln \left (x \right )+c_2 \\ \end{align*}

Substituting \(u=u \left (x \right )-\frac {2 x}{9}\) in the above solution gives

\begin{align*} \frac {\ln \left (\frac {9 u \left (x \right )-18 x}{x}\right )}{4860}+\frac {\ln \left (\frac {9 u \left (x \right )+12 x}{x}\right )}{3240}-\frac {\ln \left (\frac {9 u \left (x \right )}{x}\right )}{1944} = \frac {\ln \left (x \right )}{1944}+c_1 \end{align*}

Now we transform the solution \(u \left (x \right ) = -\frac {14 x}{9}\) to \(u \left (x \right )\) using \(u \left (x \right ) = u(x) - \frac {f_2}{3 f_3}\), which gives

\begin{align*} u \left (x \right ) &= -\frac {14 x}{9} - \left (-\frac {2 x}{9}\right ) \\ &= -\frac {4 x}{3}\\ &= -\frac {4 x}{3} \end{align*}

Now we transform the solution \(u \left (x \right ) = -\frac {2 x}{9}\) to \(u \left (x \right )\) using \(u \left (x \right ) = u(x) - \frac {f_2}{3 f_3}\), which gives

\begin{align*} u \left (x \right ) &= -\frac {2 x}{9} - \left (-\frac {2 x}{9}\right ) \\ &= 0\\ &= 0 \end{align*}

Simplifying the above gives

\begin{align*} -\frac {\ln \left (3\right )}{3240}+\frac {\ln \left (\frac {u \left (x \right )-2 x}{x}\right )}{4860}+\frac {\ln \left (\frac {3 u \left (x \right )+4 x}{x}\right )}{3240}-\frac {\ln \left (\frac {u \left (x \right )}{x}\right )}{1944} &= \frac {\ln \left (x \right )}{1944}+c_1 \\ u \left (x \right ) &= -\frac {4 x}{3} \\ u \left (x \right ) &= 0 \\ \end{align*}

Summary of solutions found

\begin{align*} -\frac {\ln \left (3\right )}{3240}+\frac {\ln \left (\frac {u \left (x \right )-2 x}{x}\right )}{4860}+\frac {\ln \left (\frac {3 u \left (x \right )+4 x}{x}\right )}{3240}-\frac {\ln \left (\frac {u \left (x \right )}{x}\right )}{1944} &= \frac {\ln \left (x \right )}{1944}+c_1 \\ u \left (x \right ) &= 0 \\ u \left (x \right ) &= -\frac {4 x}{3} \\ \end{align*}

2.1.16.21 Solved using first_order_ode_LIE

6.454 (sec)

Entering first order ode LIE solver

\begin{align*} u^{\prime }\left (x \right )&=\frac {3 u \left (x \right )^{3}}{8 x^{3}}-\frac {u \left (x \right )^{2}}{4 x^{2}} \\ \end{align*}

Writing the ode as

\begin{align*} u^{\prime }\left (x \right )&=\frac {u^{2} \left (3 u -2 x \right )}{8 x^{3}}\\ u^{\prime }\left (x \right )&= \omega \left ( x,u \left (x \right )\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

\begin{align*} \eta _{x}+\omega \left ( \eta _{u \left (x \right )}-\xi _{x}\right ) -\omega ^{2}\xi _{u \left (x \right )}-\omega _{x}\xi -\omega _{u \left (x \right )}\eta =0\tag {A} \end{align*}

To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives

\begin{align*} \tag{1E} \xi &= u a_{3}+x a_{2}+a_{1} \\ \tag{2E} \eta &= u b_{3}+x b_{2}+b_{1} \\ \end{align*}

Where the unknown coefficients are

\[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \]

Substituting equations (1E,2E) and \(\omega \) into (A) gives

\begin{equation} \tag{5E} b_{2}+\frac {u^{2} \left (3 u -2 x \right ) \left (b_{3}-a_{2}\right )}{8 x^{3}}-\frac {u^{4} \left (3 u -2 x \right )^{2} a_{3}}{64 x^{6}}-\left (-\frac {u^{2}}{4 x^{3}}-\frac {3 u^{2} \left (3 u -2 x \right )}{8 x^{4}}\right ) \left (u a_{3}+x a_{2}+a_{1}\right )-\left (\frac {u \left (3 u -2 x \right )}{4 x^{3}}+\frac {3 u^{2}}{8 x^{3}}\right ) \left (u b_{3}+x b_{2}+b_{1}\right ) = 0 \end{equation}

Putting the above in normal form gives

\[ -\frac {9 u^{6} a_{3}-12 u^{5} x a_{3}-68 u^{4} x^{2} a_{3}-48 u^{3} x^{3} a_{2}+32 u^{3} x^{3} a_{3}+48 u^{3} x^{3} b_{3}+16 u^{2} x^{4} a_{2}+72 u^{2} x^{4} b_{2}-16 u^{2} x^{4} b_{3}-32 u \,x^{5} b_{2}-64 b_{2} x^{6}-72 u^{3} x^{2} a_{1}+32 u^{2} x^{3} a_{1}+72 u^{2} x^{3} b_{1}-32 u \,x^{4} b_{1}}{64 x^{6}} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} -9 u^{6} a_{3}+12 u^{5} x a_{3}+68 u^{4} x^{2} a_{3}+48 u^{3} x^{3} a_{2}-32 u^{3} x^{3} a_{3}-48 u^{3} x^{3} b_{3}-16 u^{2} x^{4} a_{2}-72 u^{2} x^{4} b_{2}+16 u^{2} x^{4} b_{3}+32 u \,x^{5} b_{2}+64 b_{2} x^{6}+72 u^{3} x^{2} a_{1}-32 u^{2} x^{3} a_{1}-72 u^{2} x^{3} b_{1}+32 u \,x^{4} b_{1} = 0 \end{equation}

Looking at the above PDE shows the following are all the terms with \(\{u, x\}\) in them.

\[ \{u, x\} \]

The following substitution is now made to be able to collect on all terms with \(\{u, x\}\) in them

\[ \{u = v_{1}, x = v_{2}\} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} 48 a_{2} v_{1}^{3} v_{2}^{3}-16 a_{2} v_{1}^{2} v_{2}^{4}-9 a_{3} v_{1}^{6}+12 a_{3} v_{1}^{5} v_{2}+68 a_{3} v_{1}^{4} v_{2}^{2}-32 a_{3} v_{1}^{3} v_{2}^{3}-72 b_{2} v_{1}^{2} v_{2}^{4}+32 b_{2} v_{1} v_{2}^{5}+64 b_{2} v_{2}^{6}-48 b_{3} v_{1}^{3} v_{2}^{3}+16 b_{3} v_{1}^{2} v_{2}^{4}+72 a_{1} v_{1}^{3} v_{2}^{2}-32 a_{1} v_{1}^{2} v_{2}^{3}-72 b_{1} v_{1}^{2} v_{2}^{3}+32 b_{1} v_{1} v_{2}^{4} = 0 \end{equation}

Collecting the above on the terms \(v_i\) introduced, and these are

\[ \{v_{1}, v_{2}\} \]

Equation (7E) now becomes

\begin{equation} \tag{8E} -9 a_{3} v_{1}^{6}+12 a_{3} v_{1}^{5} v_{2}+68 a_{3} v_{1}^{4} v_{2}^{2}+\left (48 a_{2}-32 a_{3}-48 b_{3}\right ) v_{1}^{3} v_{2}^{3}+72 a_{1} v_{1}^{3} v_{2}^{2}+\left (-16 a_{2}-72 b_{2}+16 b_{3}\right ) v_{1}^{2} v_{2}^{4}+\left (-32 a_{1}-72 b_{1}\right ) v_{1}^{2} v_{2}^{3}+32 b_{2} v_{1} v_{2}^{5}+32 b_{1} v_{1} v_{2}^{4}+64 b_{2} v_{2}^{6} = 0 \end{equation}

Setting each coefficients in (8E) to zero gives the following equations to solve

\begin{align*} 72 a_{1}&=0\\ -9 a_{3}&=0\\ 12 a_{3}&=0\\ 68 a_{3}&=0\\ 32 b_{1}&=0\\ 32 b_{2}&=0\\ 64 b_{2}&=0\\ -32 a_{1}-72 b_{1}&=0\\ -16 a_{2}-72 b_{2}+16 b_{3}&=0\\ 48 a_{2}-32 a_{3}-48 b_{3}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=0\\ a_{2}&=b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \end{align*}

Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives

\begin{align*} \xi &= x \\ \eta &= u \\ \end{align*}

Shifting is now applied to make \(\xi =0\) in order to simplify the rest of the computation

\begin{align*} \eta &= \eta - \omega \left (x,u\right ) \xi \\ &= u - \left (\frac {u^{2} \left (3 u -2 x \right )}{8 x^{3}}\right ) \left (x\right ) \\ &= u +\frac {-3 u^{3}+2 u^{2} x}{8 x^{2}}\\ \xi &= 0 \end{align*}

The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,u\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.

The characteristic pde which is used to find the canonical coordinates is

\begin{align*} \frac {d x}{\xi } &= \frac {d u}{\eta } = dS \tag {1} \end{align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial u}\right ) S(x,u) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since \(\xi =0\) then in this special case

\begin{align*} R = x \end{align*}

\(S\) is found from

\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{u +\frac {-3 u^{3}+2 u^{2} x}{8 x^{2}}}} dy \end{align*}

Which results in

\begin{align*} S&= \ln \left (u \right )-\frac {3 \ln \left (3 u +4 x \right )}{5}-\frac {2 \ln \left (u -2 x \right )}{5} \end{align*}

Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating

\begin{align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,u) S_{u} }{ R_{x} + \omega (x,u) R_{u} }\tag {2} \end{align*}

Where in the above \(R_{x},R_{u},S_{x},S_{u}\) are all partial derivatives and \(\omega (x,u)\) is the right hand side of the original ode given by

\begin{align*} \omega (x,u) &= \frac {u^{2} \left (3 u -2 x \right )}{8 x^{3}} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= 1\\ R_{u} &= 0\\ S_{x} &= \frac {8 x}{\left (3 u +4 x \right ) \left (u -2 x \right )}\\ S_{u} &= -\frac {8 x^{2}}{u \left (3 u +4 x \right ) \left (u -2 x \right )} \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= -\frac {1}{x}\tag {2A} \end{align*}

We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,u\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives

\begin{align*} \frac {dS}{dR} &= -\frac {1}{R} \end{align*}

The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\).

Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {-\frac {1}{R}\, dR}\\ S \left (R \right ) &= -\ln \left (R \right ) + c_2 \end{align*}

\begin{align*} S \left (R \right )&= -\ln \left (R \right )+c_2 \end{align*}

To complete the solution, we just need to transform the above back to \(x,u\) coordinates. This results in

\begin{align*} \ln \left (u \left (x \right )\right )-\frac {3 \ln \left (3 u \left (x \right )+4 x \right )}{5}-\frac {2 \ln \left (u \left (x \right )-2 x \right )}{5} = -\ln \left (x \right )+c_2 \end{align*}

Summary of solutions found

\begin{align*} \ln \left (u \left (x \right )\right )-\frac {3 \ln \left (3 u \left (x \right )+4 x \right )}{5}-\frac {2 \ln \left (u \left (x \right )-2 x \right )}{5} &= -\ln \left (x \right )+c_2 \\ \end{align*}

Substituting \(u \left (x \right )=\frac {1}{y-\frac {1}{2 x}}\) in the above solution gives

\[ -\frac {\ln \left (\frac {1}{x \left (y-\frac {1}{2 x}\right )}\right )}{8}+\frac {\ln \left (\frac {\frac {1}{y-\frac {1}{2 x}}-2 x}{x}\right )}{20}+\frac {3 \ln \left (\frac {\frac {3}{y-\frac {1}{2 x}}+4 x}{x}\right )}{40} = \frac {\ln \left (x \right )}{8}+c_2 \]

Now we transform the solution \(u \left (x \right ) = 2 x\) to \(y\) using \(u \left (x \right )=\frac {1}{y-\frac {1}{2 x}}\) which gives

\[ y = \frac {1}{x} \]

Simplifying the above gives

\begin{align*} \frac {\ln \left (2\right )}{20}-\frac {\ln \left (\frac {1}{2 y x -1}\right )}{8}+\frac {\ln \left (\frac {1-y x}{2 y x -1}\right )}{20}+\frac {3 \ln \left (\frac {4 y x +1}{2 y x -1}\right )}{40} &= \frac {\ln \left (x \right )}{8}+c_2 \\ y &= \frac {1}{x} \\ \end{align*}

Direction field \(y^{\prime } = \frac {-y x -1}{4 x^{3} y-2 x^{2}}\)	Isoclines for \(y^{\prime } = \frac {-y x -1}{4 x^{3} y-2 x^{2}}\)

Summary of solutions found

\begin{align*} \frac {\ln \left (2\right )}{20}-\frac {\ln \left (\frac {1}{2 y x -1}\right )}{8}+\frac {\ln \left (\frac {1-y x}{2 y x -1}\right )}{20}+\frac {3 \ln \left (\frac {4 y x +1}{2 y x -1}\right )}{40} &= \frac {\ln \left (x \right )}{8}+c_2 \\ y &= \frac {1}{x} \\ \end{align*}

2.1.16.22 Solved using first_order_ode_LIE

11.921 (sec)

Entering first order ode LIE solver

\begin{align*} y^{\prime }&=\frac {-y x -1}{4 x^{3} y-2 x^{2}} \\ \end{align*}

Writing the ode as

\begin{align*} y^{\prime }&=-\frac {x y +1}{2 x^{2} \left (2 x y -1\right )}\\ y^{\prime }&= \omega \left ( x,y\right ) \end{align*}

The condition of Lie symmetry is the linearized PDE given by

\begin{align*} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0\tag {A} \end{align*}

To determine \(\xi ,\eta \) then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives

\begin{align*} \tag{1E} \xi &= x a_{2}+y a_{3}+a_{1} \\ \tag{2E} \eta &= x b_{2}+y b_{3}+b_{1} \\ \end{align*}

Where the unknown coefficients are

\[ \{a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}\} \]

Substituting equations (1E,2E) and \(\omega \) into (A) gives

\begin{equation} \tag{5E} b_{2}-\frac {\left (x y +1\right ) \left (b_{3}-a_{2}\right )}{2 x^{2} \left (2 x y -1\right )}-\frac {\left (x y +1\right )^{2} a_{3}}{4 x^{4} \left (2 x y -1\right )^{2}}-\left (-\frac {y}{2 x^{2} \left (2 x y -1\right )}+\frac {x y +1}{x^{3} \left (2 x y -1\right )}+\frac {\left (x y +1\right ) y}{x^{2} \left (2 x y -1\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (-\frac {1}{2 x \left (2 x y -1\right )}+\frac {x y +1}{x \left (2 x y -1\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0 \end{equation}

Putting the above in normal form gives

\[ \frac {16 x^{6} y^{2} b_{2}-16 x^{5} y b_{2}-4 x^{4} y^{2} a_{2}-4 x^{4} y^{2} b_{3}-8 x^{3} y^{3} a_{3}-8 x^{3} y^{2} a_{1}-2 b_{2} x^{4}-8 x^{3} y a_{2}-8 x^{3} y b_{3}-11 x^{2} y^{2} a_{3}-6 x^{3} b_{1}-10 x^{2} y a_{1}+2 x^{2} a_{2}+2 x^{2} b_{3}+2 x y a_{3}+4 x a_{1}-a_{3}}{4 x^{4} \left (2 x y -1\right )^{2}} = 0 \]

Setting the numerator to zero gives

\begin{equation} \tag{6E} 16 x^{6} y^{2} b_{2}-16 x^{5} y b_{2}-4 x^{4} y^{2} a_{2}-4 x^{4} y^{2} b_{3}-8 x^{3} y^{3} a_{3}-8 x^{3} y^{2} a_{1}-2 b_{2} x^{4}-8 x^{3} y a_{2}-8 x^{3} y b_{3}-11 x^{2} y^{2} a_{3}-6 x^{3} b_{1}-10 x^{2} y a_{1}+2 x^{2} a_{2}+2 x^{2} b_{3}+2 x y a_{3}+4 x a_{1}-a_{3} = 0 \end{equation}

Looking at the above PDE shows the following are all the terms with \(\{x, y\}\) in them.

\[ \{x, y\} \]

The following substitution is now made to be able to collect on all terms with \(\{x, y\}\) in them

\[ \{x = v_{1}, y = v_{2}\} \]

The above PDE (6E) now becomes

\begin{equation} \tag{7E} 16 b_{2} v_{1}^{6} v_{2}^{2}-4 a_{2} v_{1}^{4} v_{2}^{2}-8 a_{3} v_{1}^{3} v_{2}^{3}-16 b_{2} v_{1}^{5} v_{2}-4 b_{3} v_{1}^{4} v_{2}^{2}-8 a_{1} v_{1}^{3} v_{2}^{2}-8 a_{2} v_{1}^{3} v_{2}-11 a_{3} v_{1}^{2} v_{2}^{2}-2 b_{2} v_{1}^{4}-8 b_{3} v_{1}^{3} v_{2}-10 a_{1} v_{1}^{2} v_{2}-6 b_{1} v_{1}^{3}+2 a_{2} v_{1}^{2}+2 a_{3} v_{1} v_{2}+2 b_{3} v_{1}^{2}+4 a_{1} v_{1}-a_{3} = 0 \end{equation}

Collecting the above on the terms \(v_i\) introduced, and these are

\[ \{v_{1}, v_{2}\} \]

Equation (7E) now becomes

\begin{equation} \tag{8E} 16 b_{2} v_{1}^{6} v_{2}^{2}-16 b_{2} v_{1}^{5} v_{2}+\left (-4 a_{2}-4 b_{3}\right ) v_{1}^{4} v_{2}^{2}-2 b_{2} v_{1}^{4}-8 a_{3} v_{1}^{3} v_{2}^{3}-8 a_{1} v_{1}^{3} v_{2}^{2}+\left (-8 a_{2}-8 b_{3}\right ) v_{1}^{3} v_{2}-6 b_{1} v_{1}^{3}-11 a_{3} v_{1}^{2} v_{2}^{2}-10 a_{1} v_{1}^{2} v_{2}+\left (2 a_{2}+2 b_{3}\right ) v_{1}^{2}+2 a_{3} v_{1} v_{2}+4 a_{1} v_{1}-a_{3} = 0 \end{equation}

Setting each coefficients in (8E) to zero gives the following equations to solve

\begin{align*} -10 a_{1}&=0\\ -8 a_{1}&=0\\ 4 a_{1}&=0\\ -11 a_{3}&=0\\ -8 a_{3}&=0\\ -a_{3}&=0\\ 2 a_{3}&=0\\ -6 b_{1}&=0\\ -16 b_{2}&=0\\ -2 b_{2}&=0\\ 16 b_{2}&=0\\ -8 a_{2}-8 b_{3}&=0\\ -4 a_{2}-4 b_{3}&=0\\ 2 a_{2}+2 b_{3}&=0 \end{align*}

Solving the above equations for the unknowns gives

\begin{align*} a_{1}&=0\\ a_{2}&=-b_{3}\\ a_{3}&=0\\ b_{1}&=0\\ b_{2}&=0\\ b_{3}&=b_{3} \end{align*}

Substituting the above solution in the anstaz (1E,2E) (using \(1\) as arbitrary value for any unknown in the RHS) gives

\begin{align*} \xi &= -x \\ \eta &= y \\ \end{align*}

Shifting is now applied to make \(\xi =0\) in order to simplify the rest of the computation

\begin{align*} \eta &= \eta - \omega \left (x,y\right ) \xi \\ &= y - \left (-\frac {x y +1}{2 x^{2} \left (2 x y -1\right )}\right ) \left (-x\right ) \\ &= \frac {4 x^{2} y^{2}-3 x y -1}{4 x^{2} y -2 x}\\ \xi &= 0 \end{align*}

The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( x,y\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.

The characteristic pde which is used to find the canonical coordinates is

\begin{align*} \frac {d x}{\xi } &= \frac {d y}{\eta } = dS \tag {1} \end{align*}

The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial x} + \eta \frac {\partial }{\partial y}\right ) S(x,y) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since \(\xi =0\) then in this special case

\begin{align*} R = x \end{align*}

\(S\) is found from

\begin{align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{\frac {4 x^{2} y^{2}-3 x y -1}{4 x^{2} y -2 x}}} dy \end{align*}

Which results in

\begin{align*} S&= \frac {2 \ln \left (x y -1\right )}{5}+\frac {3 \ln \left (4 x y +1\right )}{5} \end{align*}

Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating

\begin{align*} \frac {dS}{dR} &= \frac { S_{x} + \omega (x,y) S_{y} }{ R_{x} + \omega (x,y) R_{y} }\tag {2} \end{align*}

Where in the above \(R_{x},R_{y},S_{x},S_{y}\) are all partial derivatives and \(\omega (x,y)\) is the right hand side of the original ode given by

\begin{align*} \omega (x,y) &= -\frac {x y +1}{2 x^{2} \left (2 x y -1\right )} \end{align*}

Evaluating all the partial derivatives gives

\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {4 x \,y^{2}-2 y}{4 x^{2} y^{2}-3 x y -1}\\ S_{y} &= \frac {4 x^{2} y -2 x}{4 x^{2} y^{2}-3 x y -1} \end{align*}

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

\begin{align*} \frac {dS}{dR} &= \frac {1}{x}\tag {2A} \end{align*}

We now need to express the RHS as function of \(R\) only. This is done by solving for \(x,y\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives

\begin{align*} \frac {dS}{dR} &= \frac {1}{R} \end{align*}

The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\).

Since the ode has the form \(\frac {d}{d R}S \left (R \right )=f(R)\), then we only need to integrate \(f(R)\).

\begin{align*} \int {dS} &= \int {\frac {1}{R}\, dR}\\ S \left (R \right ) &= \ln \left (R \right ) + c_2 \end{align*}

\begin{align*} S \left (R \right )&= \ln \left (R \right )+c_2 \end{align*}

To complete the solution, we just need to transform the above back to \(x,y\) coordinates. This results in

\begin{align*} \frac {2 \ln \left (y x -1\right )}{5}+\frac {3 \ln \left (4 y x +1\right )}{5} = \ln \left (x \right )+c_2 \end{align*}

The following diagram shows solution curves of the original ode and how they transform in the canonical coordinates space using the mapping shown.

Original ode in \(x,y\) coordinates	Canonical coordinates transformation	ODE in canonical coordinates \((R,S)\)
\( \frac {dy}{dx} = -\frac {x y +1}{2 x^{2} \left (2 x y -1\right )}\)		\( \frac {d S}{d R} = \frac {1}{R}\)
	\(\!\begin {aligned} R&= x\\ S&= \frac {2 \ln \left (x y -1\right )}{5}+\frac {3 \ln \left (4 x y +1\right )}{5} \end {aligned} \)

Direction field \(y^{\prime } = \frac {-y x -1}{4 x^{3} y-2 x^{2}}\)	Isoclines for \(y^{\prime } = \frac {-y x -1}{4 x^{3} y-2 x^{2}}\)

Summary of solutions found

\begin{align*} \frac {2 \ln \left (y x -1\right )}{5}+\frac {3 \ln \left (4 y x +1\right )}{5} &= \ln \left (x \right )+c_2 \\ \end{align*}

2.1.16.23 ✓ Maple. Time used: 0.157 (sec). Leaf size: 37

ode:=diff(y(x),x) = (-y(x)*x-1)/(4*x^3*y(x)-2*x^2); 
dsolve(ode,y(x), singsol=all);
 

\[ y = \frac {\operatorname {RootOf}\left (c_1 \,\textit {\_Z}^{25}-10 c_1 \,\textit {\_Z}^{20}+25 c_1 \,\textit {\_Z}^{15}-16 x^{5}\right )^{5}-1}{4 x} \]

Maple trace

Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
trying Bernoulli 
trying separable 
trying inverse linear 
trying homogeneous types: 
trying homogeneous G 
<- homogeneous successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {-x y \left (x \right )-1}{4 x^{3} y \left (x \right )-2 x^{2}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {-x y \left (x \right )-1}{4 x^{3} y \left (x \right )-2 x^{2}} \end {array} \]

2.1.16.24 ✓ Mathematica. Time used: 11.033 (sec). Leaf size: 391

ode=D[y[x],x] == (-x*y[x]-1)/(4*x^3*y[x]-2*x^2); 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to \text {Root}\left [64 \text {$\#$1}^5 c_1{}^5 x^5-80 \text {$\#$1}^4 c_1{}^5 x^4-20 \text {$\#$1}^3 c_1{}^5 x^3+25 \text {$\#$1}^2 c_1{}^5 x^2+10 \text {$\#$1} c_1{}^5 x-x^5+c_1{}^5\&,1\right ]\\ y(x)&\to \text {Root}\left [64 \text {$\#$1}^5 c_1{}^5 x^5-80 \text {$\#$1}^4 c_1{}^5 x^4-20 \text {$\#$1}^3 c_1{}^5 x^3+25 \text {$\#$1}^2 c_1{}^5 x^2+10 \text {$\#$1} c_1{}^5 x-x^5+c_1{}^5\&,2\right ]\\ y(x)&\to \text {Root}\left [64 \text {$\#$1}^5 c_1{}^5 x^5-80 \text {$\#$1}^4 c_1{}^5 x^4-20 \text {$\#$1}^3 c_1{}^5 x^3+25 \text {$\#$1}^2 c_1{}^5 x^2+10 \text {$\#$1} c_1{}^5 x-x^5+c_1{}^5\&,3\right ]\\ y(x)&\to \text {Root}\left [64 \text {$\#$1}^5 c_1{}^5 x^5-80 \text {$\#$1}^4 c_1{}^5 x^4-20 \text {$\#$1}^3 c_1{}^5 x^3+25 \text {$\#$1}^2 c_1{}^5 x^2+10 \text {$\#$1} c_1{}^5 x-x^5+c_1{}^5\&,4\right ]\\ y(x)&\to \text {Root}\left [64 \text {$\#$1}^5 c_1{}^5 x^5-80 \text {$\#$1}^4 c_1{}^5 x^4-20 \text {$\#$1}^3 c_1{}^5 x^3+25 \text {$\#$1}^2 c_1{}^5 x^2+10 \text {$\#$1} c_1{}^5 x-x^5+c_1{}^5\&,5\right ] \end{align*}

2.1.16.25 ✓ Sympy. Time used: 0.601 (sec). Leaf size: 29

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq((x*y(x) + 1)/(4*x**3*y(x) - 2*x**2) + Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

\[ - \log {\left (x \right )} + \frac {2 \log {\left (x y{\left (x \right )} - 1 \right )}}{5} + \frac {3 \log {\left (x y{\left (x \right )} + \frac {1}{4} \right )}}{5} = C_{1} \]

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0
 

classify_ode(ode,func=y(x)) 
 
('factorable', 'separable_reduced', 'lie_group', 'separable_reduced_Integral')