2.1.4 Internal
problem
ID
[20965]Book
:
Ordinary
Differential
Equations.
By
Wolfgang
Walter.
Graduate
texts
in
Mathematics.
Springer.
NY.
QA372.W224
1998 Section
:
Chapter
1.
First
order
equations:
Some
integrable
cases.
Excercises
XII
at
page
23 Problem
number
:
(d)Date
solved
:
Saturday, November 29, 2025 at 01:11:33 AMCAS
classification
:
[[_homogeneous, `class C`], _rational, [_Abel, `2nd type`, `class A`]]
Time used: 2.704 (sec)
Solve
\begin{align*}
y^{\prime }&=\frac {2 y+x +1}{2 x +y+2} \\
\end{align*}
Let
\(p=y^{\prime }\) the ode becomes
\begin{align*} p = \frac {2 y +x +1}{2 x +y +2} \end{align*}
Solving for \(y\) from the above results in
\begin{align*}
\tag{1} y &= -\frac {\left (2 p -1\right ) x}{p -2}-\frac {2 p -1}{p -2} \\
\end{align*}
This has the form
\begin{align*} y=x f(p)+g(p)\tag {*} \end{align*}
Where \(f,g\) are functions of \(p=y'(x)\) . The above ode is dAlembert ode which is now solved.
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 \(y=x f + g\) to (1A) shows that
\begin{align*} f &= \frac {-2 p +1}{p -2}\\ g &= \frac {-2 p +1}{p -2} \end{align*}
Hence (2) becomes
\begin{equation}
\tag{2A} p -\frac {-2 p +1}{p -2} = \left (-\frac {2 x}{p -2}+\frac {2 x p}{\left (p -2\right )^{2}}-\frac {x}{\left (p -2\right )^{2}}-\frac {2}{p -2}+\frac {2 p}{\left (p -2\right )^{2}}-\frac {1}{\left (p -2\right )^{2}}\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 {-2 p +1}{p -2} = 0 \end{align*}
Solving the above for \(p\) results in
\begin{align*} p_{1} &=1\\ p_{2} &=-1 \end{align*}
Substituting these in (1A) and keeping singular solution that verifies the ode gives
\begin{align*} y = 1+x\\ y = -1-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 {-2 p \left (x \right )+1}{p \left (x \right )-2}}{-\frac {2 x}{p \left (x \right )-2}+\frac {2 x p \left (x \right )}{\left (p \left (x \right )-2\right )^{2}}-\frac {x}{\left (p \left (x \right )-2\right )^{2}}-\frac {2}{p \left (x \right )-2}+\frac {2 p \left (x \right )}{\left (p \left (x \right )-2\right )^{2}}-\frac {1}{\left (p \left (x \right )-2\right )^{2}}}
\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 {\left (p \left (x \right )-1\right ) \left (p \left (x \right )-2\right ) \left (p \left (x \right )+1\right )}{3 x +3}
\end{equation}
is separable as it can be written as
\begin{align*} p^{\prime }\left (x \right )&= \frac {\left (p \left (x \right )-1\right ) \left (p \left (x \right )-2\right ) \left (p \left (x \right )+1\right )}{3 x +3}\\ &= f(x) g(p) \end{align*}
Where
\begin{align*} f(x) &= \frac {1}{3 x +3}\\ g(p) &= \left (p -1\right ) \left (p -2\right ) \left (p +1\right ) \end{align*}
Integrating gives
\begin{align*}
\int { \frac {1}{g(p)} \,dp} &= \int { f(x) \,dx} \\
\int { \frac {1}{\left (p -1\right ) \left (p -2\right ) \left (p +1\right )}\,dp} &= \int { \frac {1}{3 x +3} \,dx} \\
\end{align*}
\[
\frac {\ln \left (p \left (x \right )+1\right )}{6}+\frac {\ln \left (p \left (x \right )-2\right )}{3}-\frac {\ln \left (p \left (x \right )-1\right )}{2}=\ln \left (\left (1+x \right )^{{1}/{3}}\right )+c_1
\]
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
\[
\left (p -1\right ) \left (p -2\right ) \left (p +1\right )=0
\]
for
\(p \left (x \right )\) gives
\begin{align*} p \left (x \right )&=-1\\ p \left (x \right )&=1 \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 (p \left (x \right )+1\right )}{6}+\frac {\ln \left (p \left (x \right )-2\right )}{3}-\frac {\ln \left (p \left (x \right )-1\right )}{2} &= \ln \left (\left (1+x \right )^{{1}/{3}}\right )+c_1 \\
p \left (x \right ) &= -1 \\
p \left (x \right ) &= 1 \\
\end{align*}
Substituing the above solution for
\(p\) in (2A) gives
\begin{align*}
\text {Expression too large to display} \\
y &= -1-x \\
y &= 1+x \\
\end{align*}
Simplifying
the above gives
\begin{align*}
y &= 1+x \\
y &= -1-x \\
y &= -\frac {2 \left (1+x \right ) \left (1+\frac {\left (-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}\right ) {\left (\left (-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}\right )^{2} \left ({\mathrm e}^{3 c_1} \sqrt {\frac {\left (1+x \right )^{2}}{-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}}}+1\right )\right )}^{{1}/{3}}}{2}-{\mathrm e}^{6 c_1} \left (1+x \right )^{2}+{\left (\left (-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}\right )^{2} \left ({\mathrm e}^{3 c_1} \sqrt {\frac {\left (1+x \right )^{2}}{-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}}}+1\right )\right )}^{{2}/{3}}\right )}{\left (1-{\mathrm e}^{6 c_1} \left (1+x \right )^{2}\right ) {\left (\left (-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}\right )^{2} \left ({\mathrm e}^{3 c_1} \sqrt {\frac {\left (1+x \right )^{2}}{-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}}}+1\right )\right )}^{{1}/{3}}-{\mathrm e}^{6 c_1} \left (1+x \right )^{2}+{\left (\left (-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}\right )^{2} \left ({\mathrm e}^{3 c_1} \sqrt {\frac {\left (1+x \right )^{2}}{-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}}}+1\right )\right )}^{{2}/{3}}+1} \\
y &= -1-x \\
y &= 1+x \\
\end{align*}
Figure 2.8: Slope field \(y^{\prime } = \frac {2 y+x +1}{2 x +y+2}\) Summary of solutions found
\begin{align*}
y &= -\frac {2 \left (1+x \right ) \left (1+\frac {\left (-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}\right ) {\left (\left (-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}\right )^{2} \left ({\mathrm e}^{3 c_1} \sqrt {\frac {\left (1+x \right )^{2}}{-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}}}+1\right )\right )}^{{1}/{3}}}{2}-{\mathrm e}^{6 c_1} \left (1+x \right )^{2}+{\left (\left (-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}\right )^{2} \left ({\mathrm e}^{3 c_1} \sqrt {\frac {\left (1+x \right )^{2}}{-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}}}+1\right )\right )}^{{2}/{3}}\right )}{\left (1-{\mathrm e}^{6 c_1} \left (1+x \right )^{2}\right ) {\left (\left (-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}\right )^{2} \left ({\mathrm e}^{3 c_1} \sqrt {\frac {\left (1+x \right )^{2}}{-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}}}+1\right )\right )}^{{1}/{3}}-{\mathrm e}^{6 c_1} \left (1+x \right )^{2}+{\left (\left (-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}\right )^{2} \left ({\mathrm e}^{3 c_1} \sqrt {\frac {\left (1+x \right )^{2}}{-1+{\mathrm e}^{6 c_1} \left (1+x \right )^{2}}}+1\right )\right )}^{{2}/{3}}+1} \\
y &= -1-x \\
y &= 1+x \\
\end{align*}
Time used: 1.125 (sec)
Solve
\begin{align*}
y^{\prime }&=\frac {2 y+x +1}{2 x +y+2} \\
\end{align*}
Let
\(Y = y -y_{0}\) and
\(X = x -x_{0}\) then the above is transformed to new ode in
\(Y(X)\) \[
\frac {d}{d X}Y \left (X \right ) = \frac {2 Y \left (X \right )+2 y_{0} +X +x_{0} +1}{2 X +2 x_{0} +Y \left (X \right )+y_{0} +2}
\]
Solving for possible values of
\(x_{0}\) and
\(y_{0}\)
which makes the above ode a homogeneous ode results in
\begin{align*} x_{0}&=-1\\ 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 )+X}{2 X +Y \left (X \right )} \end{align*}
In canonical form, the ODE is
\begin{align*} Y' &= F(X,Y)\\ &= \frac {2 Y +X}{2 X +Y}\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
\(M=2 Y +X\) and
\(N=2 X +Y\) are both homogeneous and of the same order
\(n=1\) . 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 {2 u +1}{u +2}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}X}} &= \frac {\frac {2 u \left (X \right )+1}{u \left (X \right )+2}-u \left (X \right )}{X} \end{align*}
Or
\[ \frac {d}{d X}u \left (X \right )-\frac {\frac {2 u \left (X \right )+1}{u \left (X \right )+2}-u \left (X \right )}{X} = 0 \]
Or
\[ \left (\frac {d}{d X}u \left (X \right )\right ) X u \left (X \right )+u \left (X \right )^{2}+2 \left (\frac {d}{d X}u \left (X \right )\right ) X -1 = 0 \]
Or
\[ \left (u \left (X \right )+2\right ) X \left (\frac {d}{d X}u \left (X \right )\right )+u \left (X \right )^{2}-1 = 0 \]
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 (u \left (X \right )-1\right ) \left (u \left (X \right )+1\right )}{\left (u \left (X \right )+2\right ) X}
\end{equation}
is separable as it can be written as
\begin{align*} \frac {d}{d X}u \left (X \right )&= -\frac {\left (u \left (X \right )-1\right ) \left (u \left (X \right )+1\right )}{\left (u \left (X \right )+2\right ) X}\\ &= f(X) g(u) \end{align*}
Where
\begin{align*} f(X) &= -\frac {1}{X}\\ g(u) &= \frac {\left (u -1\right ) \left (u +1\right )}{u +2} \end{align*}
Integrating gives
\begin{align*}
\int { \frac {1}{g(u)} \,du} &= \int { f(X) \,dX} \\
\int { \frac {u +2}{\left (u -1\right ) \left (u +1\right )}\,du} &= \int { -\frac {1}{X} \,dX} \\
\end{align*}
\[
-\frac {\ln \left (u \left (X \right )+1\right )}{2}+\frac {3 \ln \left (u \left (X \right )-1\right )}{2}=\ln \left (\frac {1}{X}\right )+c_1
\]
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
\[
\frac {\left (u -1\right ) \left (u +1\right )}{u +2}=0
\]
for
\(u \left (X \right )\) gives
\begin{align*} u \left (X \right )&=-1\\ u \left (X \right )&=1 \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 )+1\right )}{2}+\frac {3 \ln \left (u \left (X \right )-1\right )}{2} &= \ln \left (\frac {1}{X}\right )+c_1 \\
u \left (X \right ) &= -1 \\
u \left (X \right ) &= 1 \\
\end{align*}
Converting
\(-\frac {\ln \left (u \left (X \right )+1\right )}{2}+\frac {3 \ln \left (u \left (X \right )-1\right )}{2} = \ln \left (\frac {1}{X}\right )+c_1\) back to
\(Y \left (X \right )\) gives
\begin{align*} -\frac {\ln \left (\frac {Y \left (X \right )+X}{X}\right )}{2}+\frac {3 \ln \left (\frac {Y \left (X \right )-X}{X}\right )}{2} = \ln \left (\frac {1}{X}\right )+c_1 \end{align*}
Converting \(u \left (X \right ) = -1\) back to \(Y \left (X \right )\) gives
\begin{align*} Y \left (X \right ) = -X \end{align*}
Converting \(u \left (X \right ) = 1\) back to \(Y \left (X \right )\) gives
\begin{align*} Y \left (X \right ) = X \end{align*}
Using the solution for \(Y(X)\)
\begin{align*} -\frac {\ln \left (\frac {Y \left (X \right )+X}{X}\right )}{2}+\frac {3 \ln \left (\frac {Y \left (X \right )-X}{X}\right )}{2} = \ln \left (\frac {1}{X}\right )+c_1\tag {A} \end{align*}
And replacing back terms in the above solution using
\begin{align*} Y &= y +y_{0}\\ X &= x_{0} +x \end{align*}
Or
\begin{align*} Y &= y\\ X &= x -1 \end{align*}
Then the solution in \(y\) becomes using EQ (A)
\begin{align*} -\frac {\ln \left (\frac {x +y+1}{1+x}\right )}{2}+\frac {3 \ln \left (\frac {y-x -1}{1+x}\right )}{2} = \ln \left (\frac {1}{1+x}\right )+c_1 \end{align*}
Using the solution for \(Y(X)\)
\begin{align*} Y \left (X \right ) = -X\tag {A} \end{align*}
And replacing back terms in the above solution using
\begin{align*} Y &= y +y_{0}\\ X &= x_{0} +x \end{align*}
Or
\begin{align*} Y &= y\\ X &= x -1 \end{align*}
Then the solution in \(y\) becomes using EQ (A)
\begin{align*} y = -1-x \end{align*}
Using the solution for \(Y(X)\)
\begin{align*} Y \left (X \right ) = X\tag {A} \end{align*}
And replacing back terms in the above solution using
\begin{align*} Y &= y +y_{0}\\ X &= x_{0} +x \end{align*}
Or
\begin{align*} Y &= y\\ X &= x -1 \end{align*}
Then the solution in \(y\) becomes using EQ (A)
\begin{align*} y = 1+x \end{align*}
Figure 2.9: Slope field \(y^{\prime } = \frac {2 y+x +1}{2 x +y+2}\) Summary of solutions found
\begin{align*}
-\frac {\ln \left (\frac {x +y+1}{1+x}\right )}{2}+\frac {3 \ln \left (\frac {y-x -1}{1+x}\right )}{2} &= \ln \left (\frac {1}{1+x}\right )+c_1 \\
y &= -1-x \\
y &= 1+x \\
\end{align*}
Solve
\begin{align*}
y^{\prime }&=\frac {2 y+x +1}{2 x +y+2} \\
\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 \left (x \right )+x u^{\prime }\left (x \right ) = \frac {2 x u \left (x \right )+x +1}{2 x +x u \left (x \right )+2} \end{align*}
Which is now solved Unknown ode type.
Time used: 25.183 (sec)
This is Abel second kind ODE, it has the form
\[ \left (y \left (x \right )+g\right )y^{\prime }\left (x \right )= f_0(x)+f_1(x) y \left (x \right ) +f_2(x)y \left (x \right )^{2}+f_3(x)y \left (x \right )^{3} \]
Comparing the above to given ODE which is
\begin{align*}y^{\prime }\left (x \right ) = \frac {2 y \left (x \right )+x +1}{2 x +y \left (x \right )+2}\tag {1} \end{align*}
Shows that
\begin{align*} g &= 2 x +2\\ f_0 &= 1+x\\ f_1 &= 2\\ f_2 &= 0\\ f_3 &= 0 \end{align*}
Applying transformation
\begin{align*} y \left (x \right )&=\frac {1}{u(x)}-g \end{align*}
Results in the new ode which is Abel first kind
\begin{align*} u^{\prime }\left (x \right ) = \left (3 x +3\right ) u \left (x \right )^{3}-4 u \left (x \right )^{2} \end{align*}
Which is now solved.
Solve 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 )&=\left (3 x +3\right ) u \left (x \right )^{3}-4 u \left (x \right )^{2}\tag {1} \end{align*}
Therefore
\begin{align*} f_0 &= 0\\ f_1 &= 0\\ f_2 &= -4\\ f_3 &= 3 x +3 \end{align*}
Hence
\begin{align*} f'_{0} &= 0\\ f'_{3} &= 3 \end{align*}
Since \(f_2(x)=-4\) 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 {-4}{9 x +9} \right ) \\ &= u \left (x \right )+\frac {4}{9 x +9} \end{align*}
The above transformation applied to (1) gives a new ODE as
\begin{align*} u^{\prime }\left (x \right ) = \frac {\left (729 x^{3}+2187 x^{2}+2187 x +729\right ) u \left (x \right )^{3}}{243 \left (1+x \right )^{2}}+\frac {\left (-432 x -432\right ) u \left (x \right )}{243 \left (1+x \right )^{2}}-\frac {20}{243 \left (1+x \right )^{2}}\tag {2} \end{align*}
The above ODE (2) can now be solved.
Solve Writing the ode as
\begin{align*} u^{\prime }\left (x \right )&=\frac {729 u^{3} x^{3}+2187 u^{3} x^{2}+2187 u^{3} x +729 u^{3}-432 u x -432 u -20}{243 \left (1+x \right )^{2}}\\ 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} x^{3}+2187 u^{3} x^{2}+2187 u^{3} x +729 u^{3}-432 u x -432 u -20\right ) \left (b_{3}-a_{2}\right )}{243 \left (1+x \right )^{2}}-\frac {\left (729 u^{3} x^{3}+2187 u^{3} x^{2}+2187 u^{3} x +729 u^{3}-432 u x -432 u -20\right )^{2} a_{3}}{59049 \left (1+x \right )^{4}}-\left (\frac {2187 u^{3} x^{2}+4374 u^{3} x +2187 u^{3}-432 u}{243 \left (1+x \right )^{2}}-\frac {2 \left (729 u^{3} x^{3}+2187 u^{3} x^{2}+2187 u^{3} x +729 u^{3}-432 u x -432 u -20\right )}{243 \left (1+x \right )^{3}}\right ) \left (u a_{3}+x a_{2}+a_{1}\right )-\frac {\left (2187 u^{2} x^{3}+6561 u^{2} x^{2}+6561 u^{2} x +2187 u^{2}-432 x -432\right ) \left (u b_{3}+x b_{2}+b_{1}\right )}{243 \left (1+x \right )^{2}} = 0
\end{equation}
Putting the above in normal form gives
\[
-\frac {531441 u^{6} x^{6} a_{3}+3188646 u^{6} x^{5} a_{3}+7971615 u^{6} x^{4} a_{3}+10628820 u^{6} x^{3} a_{3}+7971615 u^{6} x^{2} a_{3}-452709 u^{4} x^{4} a_{3}+354294 u^{3} x^{5} a_{2}+354294 u^{3} x^{5} b_{3}+531441 u^{2} x^{6} b_{2}+3188646 u^{6} x a_{3}-1810836 u^{4} x^{3} a_{3}+177147 u^{3} x^{4} a_{1}+1594323 u^{3} x^{4} a_{2}+1771470 u^{3} x^{4} b_{3}+531441 u^{2} x^{5} b_{1}+2657205 u^{2} x^{5} b_{2}+531441 u^{6} a_{3}-2716254 u^{4} x^{2} a_{3}+708588 u^{3} x^{3} a_{1}+2834352 u^{3} x^{3} a_{2}-29160 u^{3} x^{3} a_{3}+3542940 u^{3} x^{3} b_{3}+2657205 u^{2} x^{4} b_{1}+5314410 u^{2} x^{4} b_{2}-1810836 u^{4} x a_{3}+1062882 u^{3} x^{2} a_{1}+2480058 u^{3} x^{2} a_{2}-87480 u^{3} x^{2} a_{3}+3542940 u^{3} x^{2} b_{3}+5314410 u^{2} x^{3} b_{1}+5314410 u^{2} x^{3} b_{2}-452709 u^{4} a_{3}+708588 u^{3} x a_{1}+1062882 u^{3} x a_{2}-87480 u^{3} x a_{3}+1771470 u^{3} x b_{3}+291600 u^{2} x^{2} a_{3}+5314410 u^{2} x^{2} b_{1}+2657205 u^{2} x^{2} b_{2}-164025 x^{4} b_{2}+177147 u^{3} a_{1}+177147 u^{3} a_{2}-29160 u^{3} a_{3}+354294 u^{3} b_{3}+583200 u^{2} x a_{3}+2657205 u^{2} x b_{1}+531441 u^{2} x b_{2}+104976 u \,x^{2} a_{1}-104976 u \,x^{2} a_{2}-104976 x^{3} b_{1}-551124 x^{3} b_{2}+291600 u^{2} a_{3}+531441 u^{2} b_{1}+209952 u x a_{1}-209952 u x a_{2}+27000 u x a_{3}+4860 x^{2} a_{2}-314928 x^{2} b_{1}-669222 x^{2} b_{2}+4860 x^{2} b_{3}+104976 u a_{1}-104976 u a_{2}+27000 u a_{3}+9720 x a_{1}-314928 x b_{1}-341172 x b_{2}+9720 x b_{3}+9720 a_{1}-4860 a_{2}+400 a_{3}-104976 b_{1}-59049 b_{2}+4860 b_{3}}{59049 \left (1+x \right )^{4}} = 0
\]
Setting the numerator to zero gives
\begin{equation}
\tag{6E} -531441 u^{6} x^{6} a_{3}-3188646 u^{6} x^{5} a_{3}-7971615 u^{6} x^{4} a_{3}-10628820 u^{6} x^{3} a_{3}-7971615 u^{6} x^{2} a_{3}+452709 u^{4} x^{4} a_{3}-354294 u^{3} x^{5} a_{2}-354294 u^{3} x^{5} b_{3}-531441 u^{2} x^{6} b_{2}-3188646 u^{6} x a_{3}+1810836 u^{4} x^{3} a_{3}-177147 u^{3} x^{4} a_{1}-1594323 u^{3} x^{4} a_{2}-1771470 u^{3} x^{4} b_{3}-531441 u^{2} x^{5} b_{1}-2657205 u^{2} x^{5} b_{2}-531441 u^{6} a_{3}+2716254 u^{4} x^{2} a_{3}-708588 u^{3} x^{3} a_{1}-2834352 u^{3} x^{3} a_{2}+29160 u^{3} x^{3} a_{3}-3542940 u^{3} x^{3} b_{3}-2657205 u^{2} x^{4} b_{1}-5314410 u^{2} x^{4} b_{2}+1810836 u^{4} x a_{3}-1062882 u^{3} x^{2} a_{1}-2480058 u^{3} x^{2} a_{2}+87480 u^{3} x^{2} a_{3}-3542940 u^{3} x^{2} b_{3}-5314410 u^{2} x^{3} b_{1}-5314410 u^{2} x^{3} b_{2}+452709 u^{4} a_{3}-708588 u^{3} x a_{1}-1062882 u^{3} x a_{2}+87480 u^{3} x a_{3}-1771470 u^{3} x b_{3}-291600 u^{2} x^{2} a_{3}-5314410 u^{2} x^{2} b_{1}-2657205 u^{2} x^{2} b_{2}+164025 x^{4} b_{2}-177147 u^{3} a_{1}-177147 u^{3} a_{2}+29160 u^{3} a_{3}-354294 u^{3} b_{3}-583200 u^{2} x a_{3}-2657205 u^{2} x b_{1}-531441 u^{2} x b_{2}-104976 u \,x^{2} a_{1}+104976 u \,x^{2} a_{2}+104976 x^{3} b_{1}+551124 x^{3} b_{2}-291600 u^{2} a_{3}-531441 u^{2} b_{1}-209952 u x a_{1}+209952 u x a_{2}-27000 u x a_{3}-4860 x^{2} a_{2}+314928 x^{2} b_{1}+669222 x^{2} b_{2}-4860 x^{2} b_{3}-104976 u a_{1}+104976 u a_{2}-27000 u a_{3}-9720 x a_{1}+314928 x b_{1}+341172 x b_{2}-9720 x b_{3}-9720 a_{1}+4860 a_{2}-400 a_{3}+104976 b_{1}+59049 b_{2}-4860 b_{3} = 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} -531441 a_{3} v_{1}^{6} v_{2}^{6}-3188646 a_{3} v_{1}^{6} v_{2}^{5}-7971615 a_{3} v_{1}^{6} v_{2}^{4}-10628820 a_{3} v_{1}^{6} v_{2}^{3}-354294 a_{2} v_{1}^{3} v_{2}^{5}-7971615 a_{3} v_{1}^{6} v_{2}^{2}+452709 a_{3} v_{1}^{4} v_{2}^{4}-531441 b_{2} v_{1}^{2} v_{2}^{6}-354294 b_{3} v_{1}^{3} v_{2}^{5}-177147 a_{1} v_{1}^{3} v_{2}^{4}-1594323 a_{2} v_{1}^{3} v_{2}^{4}-3188646 a_{3} v_{1}^{6} v_{2}+1810836 a_{3} v_{1}^{4} v_{2}^{3}-531441 b_{1} v_{1}^{2} v_{2}^{5}-2657205 b_{2} v_{1}^{2} v_{2}^{5}-1771470 b_{3} v_{1}^{3} v_{2}^{4}-708588 a_{1} v_{1}^{3} v_{2}^{3}-2834352 a_{2} v_{1}^{3} v_{2}^{3}-531441 a_{3} v_{1}^{6}+2716254 a_{3} v_{1}^{4} v_{2}^{2}+29160 a_{3} v_{1}^{3} v_{2}^{3}-2657205 b_{1} v_{1}^{2} v_{2}^{4}-5314410 b_{2} v_{1}^{2} v_{2}^{4}-3542940 b_{3} v_{1}^{3} v_{2}^{3}-1062882 a_{1} v_{1}^{3} v_{2}^{2}-2480058 a_{2} v_{1}^{3} v_{2}^{2}+1810836 a_{3} v_{1}^{4} v_{2}+87480 a_{3} v_{1}^{3} v_{2}^{2}-5314410 b_{1} v_{1}^{2} v_{2}^{3}-5314410 b_{2} v_{1}^{2} v_{2}^{3}-3542940 b_{3} v_{1}^{3} v_{2}^{2}-708588 a_{1} v_{1}^{3} v_{2}-1062882 a_{2} v_{1}^{3} v_{2}+452709 a_{3} v_{1}^{4}+87480 a_{3} v_{1}^{3} v_{2}-291600 a_{3} v_{1}^{2} v_{2}^{2}-5314410 b_{1} v_{1}^{2} v_{2}^{2}-2657205 b_{2} v_{1}^{2} v_{2}^{2}+164025 b_{2} v_{2}^{4}-1771470 b_{3} v_{1}^{3} v_{2}-177147 a_{1} v_{1}^{3}-104976 a_{1} v_{1} v_{2}^{2}-177147 a_{2} v_{1}^{3}+104976 a_{2} v_{1} v_{2}^{2}+29160 a_{3} v_{1}^{3}-583200 a_{3} v_{1}^{2} v_{2}-2657205 b_{1} v_{1}^{2} v_{2}+104976 b_{1} v_{2}^{3}-531441 b_{2} v_{1}^{2} v_{2}+551124 b_{2} v_{2}^{3}-354294 b_{3} v_{1}^{3}-209952 a_{1} v_{1} v_{2}+209952 a_{2} v_{1} v_{2}-4860 a_{2} v_{2}^{2}-291600 a_{3} v_{1}^{2}-27000 a_{3} v_{1} v_{2}-531441 b_{1} v_{1}^{2}+314928 b_{1} v_{2}^{2}+669222 b_{2} v_{2}^{2}-4860 b_{3} v_{2}^{2}-104976 a_{1} v_{1}-9720 a_{1} v_{2}+104976 a_{2} v_{1}-27000 a_{3} v_{1}+314928 b_{1} v_{2}+341172 b_{2} v_{2}-9720 b_{3} v_{2}-9720 a_{1}+4860 a_{2}-400 a_{3}+104976 b_{1}+59049 b_{2}-4860 b_{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} \left (-354294 a_{2}-354294 b_{3}\right ) v_{1}^{3} v_{2}^{5}+\left (-177147 a_{1}-1594323 a_{2}-1771470 b_{3}\right ) v_{1}^{3} v_{2}^{4}+\left (-708588 a_{1}-2834352 a_{2}+29160 a_{3}-3542940 b_{3}\right ) v_{1}^{3} v_{2}^{3}+\left (-1062882 a_{1}-2480058 a_{2}+87480 a_{3}-3542940 b_{3}\right ) v_{1}^{3} v_{2}^{2}+\left (-708588 a_{1}-1062882 a_{2}+87480 a_{3}-1771470 b_{3}\right ) v_{1}^{3} v_{2}+\left (-531441 b_{1}-2657205 b_{2}\right ) v_{1}^{2} v_{2}^{5}+\left (-2657205 b_{1}-5314410 b_{2}\right ) v_{1}^{2} v_{2}^{4}+\left (-5314410 b_{1}-5314410 b_{2}\right ) v_{1}^{2} v_{2}^{3}+\left (-291600 a_{3}-5314410 b_{1}-2657205 b_{2}\right ) v_{1}^{2} v_{2}^{2}+\left (-583200 a_{3}-2657205 b_{1}-531441 b_{2}\right ) v_{1}^{2} v_{2}+\left (-104976 a_{1}+104976 a_{2}\right ) v_{1} v_{2}^{2}+\left (-209952 a_{1}+209952 a_{2}-27000 a_{3}\right ) v_{1} v_{2}+\left (-177147 a_{1}-177147 a_{2}+29160 a_{3}-354294 b_{3}\right ) v_{1}^{3}+\left (-291600 a_{3}-531441 b_{1}\right ) v_{1}^{2}+\left (-104976 a_{1}+104976 a_{2}-27000 a_{3}\right ) v_{1}+\left (104976 b_{1}+551124 b_{2}\right ) v_{2}^{3}+\left (-4860 a_{2}+314928 b_{1}+669222 b_{2}-4860 b_{3}\right ) v_{2}^{2}+\left (-9720 a_{1}+314928 b_{1}+341172 b_{2}-9720 b_{3}\right ) v_{2}-531441 a_{3} v_{1}^{6}+452709 a_{3} v_{1}^{4}+164025 b_{2} v_{2}^{4}+1810836 a_{3} v_{1}^{4} v_{2}-531441 a_{3} v_{1}^{6} v_{2}^{6}-3188646 a_{3} v_{1}^{6} v_{2}^{5}-7971615 a_{3} v_{1}^{6} v_{2}^{4}-10628820 a_{3} v_{1}^{6} v_{2}^{3}-7971615 a_{3} v_{1}^{6} v_{2}^{2}+452709 a_{3} v_{1}^{4} v_{2}^{4}-531441 b_{2} v_{1}^{2} v_{2}^{6}-3188646 a_{3} v_{1}^{6} v_{2}+1810836 a_{3} v_{1}^{4} v_{2}^{3}+2716254 a_{3} v_{1}^{4} v_{2}^{2}-9720 a_{1}+4860 a_{2}-400 a_{3}+104976 b_{1}+59049 b_{2}-4860 b_{3} = 0
\end{equation}
Setting each coefficients in (8E) to zero gives the following equations to solve
\begin{align*} -10628820 a_{3}&=0\\ -7971615 a_{3}&=0\\ -3188646 a_{3}&=0\\ -531441 a_{3}&=0\\ 452709 a_{3}&=0\\ 1810836 a_{3}&=0\\ 2716254 a_{3}&=0\\ -531441 b_{2}&=0\\ 164025 b_{2}&=0\\ -104976 a_{1}+104976 a_{2}&=0\\ -354294 a_{2}-354294 b_{3}&=0\\ -291600 a_{3}-531441 b_{1}&=0\\ -5314410 b_{1}-5314410 b_{2}&=0\\ -2657205 b_{1}-5314410 b_{2}&=0\\ -531441 b_{1}-2657205 b_{2}&=0\\ 104976 b_{1}+551124 b_{2}&=0\\ -209952 a_{1}+209952 a_{2}-27000 a_{3}&=0\\ -177147 a_{1}-1594323 a_{2}-1771470 b_{3}&=0\\ -104976 a_{1}+104976 a_{2}-27000 a_{3}&=0\\ -583200 a_{3}-2657205 b_{1}-531441 b_{2}&=0\\ -291600 a_{3}-5314410 b_{1}-2657205 b_{2}&=0\\ -1062882 a_{1}-2480058 a_{2}+87480 a_{3}-3542940 b_{3}&=0\\ -708588 a_{1}-2834352 a_{2}+29160 a_{3}-3542940 b_{3}&=0\\ -708588 a_{1}-1062882 a_{2}+87480 a_{3}-1771470 b_{3}&=0\\ -177147 a_{1}-177147 a_{2}+29160 a_{3}-354294 b_{3}&=0\\ -9720 a_{1}+314928 b_{1}+341172 b_{2}-9720 b_{3}&=0\\ -4860 a_{2}+314928 b_{1}+669222 b_{2}-4860 b_{3}&=0\\ -9720 a_{1}+4860 a_{2}-400 a_{3}+104976 b_{1}+59049 b_{2}-4860 b_{3}&=0 \end{align*}
Solving the above equations for the unknowns gives
\begin{align*} a_{1}&=-b_{3}\\ 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 &= -1-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} x^{3}+2187 u^{3} x^{2}+2187 u^{3} x +729 u^{3}-432 u x -432 u -20}{243 \left (1+x \right )^{2}}\right ) \left (-1-x\right ) \\ &= \frac {-20+729 \left (1+x \right )^{3} u^{3}+\left (-189 x -189\right ) u}{243+243 x}\\ \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 {-20+729 \left (1+x \right )^{3} u^{3}+\left (-189 x -189\right ) u}{243+243 x}}} dy \end{align*}
Which results in
\begin{align*} S&= \left (243+243 x \right ) \left (-\frac {\ln \left (9 u x +9 u +1\right )}{18 \left (9 x +9\right )}+\frac {\ln \left (9 u x +9 u +4\right )}{243+243 x}+\frac {\ln \left (9 u x +9 u -5\right )}{486 x +486}\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} x^{3}+2187 u^{3} x^{2}+2187 u^{3} x +729 u^{3}-432 u x -432 u -20}{243 \left (1+x \right )^{2}} \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= 1\\ R_{u} &= 0\\ S_{x} &= \frac {243 u}{-20+\left (9 x +9\right )^{3} u^{3}+\left (-189 x -189\right ) u}\\ S_{u} &= \frac {243+243 x}{-20+\left (9 x +9\right )^{3} u^{3}+\left (-189 x -189\right ) u} \end{align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin{align*} \frac {dS}{dR} &= \frac {1}{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 +1} \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 +1}\, dR}\\ S \left (R \right ) &= \ln \left (R +1\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 (1+u \left (x \right ) \left (9 x +9\right )\right )}{2}+\ln \left (u \left (x \right ) \left (9 x +9\right )+4\right )+\frac {\ln \left (-5+u \left (x \right ) \left (9 x +9\right )\right )}{2} = \ln \left (1+x \right )+c_2 \end{align*}
Substituting \(u=u \left (x \right )-\frac {4}{3 \left (3 x +3\right )}\) in the above solution gives
\begin{align*} -\frac {3 \ln \left (1+\left (u \left (x \right )-\frac {4}{3 \left (3 x +3\right )}\right ) \left (9 x +9\right )\right )}{2}+\ln \left (\left (u \left (x \right )-\frac {4}{3 \left (3 x +3\right )}\right ) \left (9 x +9\right )+4\right )+\frac {\ln \left (-5+\left (u \left (x \right )-\frac {4}{3 \left (3 x +3\right )}\right ) \left (9 x +9\right )\right )}{2} = \ln \left (x +1\right )+c_2 \end{align*}
Simplifying the above gives
\begin{align*}
\frac {3 \ln \left (3\right )}{2}-\frac {3 \ln \left (-1+\left (3 x +3\right ) u \left (x \right )\right )}{2}+\ln \left (u \left (x \right ) \left (x +1\right )\right )+\frac {\ln \left (u \left (x \right ) x +u \left (x \right )-1\right )}{2} &= \ln \left (x +1\right )+c_2 \\
\end{align*}
Substituting
\(u \left (x \right )=\frac {1}{2 x +y \left (x \right )+2}\) in the above solution gives
\[
\frac {3 \ln \left (3\right )}{2}-\frac {3 \ln \left (-1+\frac {3 x +3}{2 x +y \left (x \right )+2}\right )}{2}+\ln \left (\frac {x +1}{2 x +y \left (x \right )+2}\right )+\frac {\ln \left (\frac {x}{2 x +y \left (x \right )+2}+\frac {1}{2 x +y \left (x \right )+2}-1\right )}{2} = \ln \left (x +1\right )+c_2
\]
Simplifying the above gives
\begin{align*}
\frac {3 \ln \left (3\right )}{2}-\frac {3 \ln \left (\frac {-y \left (x \right )+x +1}{2 x +y \left (x \right )+2}\right )}{2}+\ln \left (\frac {x +1}{2 x +y \left (x \right )+2}\right )+\frac {\ln \left (\frac {-x -y \left (x \right )-1}{2 x +y \left (x \right )+2}\right )}{2} &= \ln \left (x +1\right )+c_2 \\
\end{align*}
Figure 2.10: Slope field \(y^{\prime }\left (x \right ) = \frac {2 y \left (x \right )+x +1}{2 x +y \left (x \right )+2}\) Figure 2.11: Slope field \(y^{\prime }\left (x \right ) = \frac {2 y \left (x \right )+x +1}{2 x +y \left (x \right )+2}\) Summary of solutions found
\begin{align*}
\frac {3 \ln \left (3\right )}{2}-\frac {3 \ln \left (\frac {-y \left (x \right )+x +1}{2 x +y \left (x \right )+2}\right )}{2}+\ln \left (\frac {x +1}{2 x +y \left (x \right )+2}\right )+\frac {\ln \left (\frac {-x -y \left (x \right )-1}{2 x +y \left (x \right )+2}\right )}{2} &= \ln \left (x +1\right )+c_2 \\
\end{align*}
Time used: 16.804 (sec)
Solve
\begin{align*}
y^{\prime }\left (x \right )&=\frac {2 y \left (x \right )+x +1}{2 x +y \left (x \right )+2} \\
\end{align*}
Writing the ode as
\begin{align*} y^{\prime }\left (x \right )&=\frac {2 y +x +1}{2 x +y +2}\\ y^{\prime }\left (x \right )&= \omega \left ( x,y \left (x \right )\right ) \end{align*}
The condition of Lie symmetry is the linearized PDE given by
\begin{align*} \eta _{x}+\omega \left ( \eta _{y \left (x \right )}-\xi _{x}\right ) -\omega ^{2}\xi _{y \left (x \right )}-\omega _{x}\xi -\omega _{y \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 &= 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 (2 y +x +1\right ) \left (b_{3}-a_{2}\right )}{2 x +y +2}-\frac {\left (2 y +x +1\right )^{2} a_{3}}{\left (2 x +y +2\right )^{2}}-\left (\frac {1}{2 x +y +2}-\frac {2 \left (2 y +x +1\right )}{\left (2 x +y +2\right )^{2}}\right ) \left (x a_{2}+y a_{3}+a_{1}\right )-\left (\frac {2}{2 x +y +2}-\frac {2 y +x +1}{\left (2 x +y +2\right )^{2}}\right ) \left (x b_{2}+y b_{3}+b_{1}\right ) = 0
\end{equation}
Putting the above in normal form gives
\[
-\frac {2 x^{2} a_{2}+x^{2} a_{3}-x^{2} b_{2}-2 x^{2} b_{3}+2 x y a_{2}+4 x y a_{3}-4 x y b_{2}-2 x y b_{3}+2 y^{2} a_{2}+y^{2} a_{3}-y^{2} b_{2}-2 y^{2} b_{3}+4 x a_{2}+2 x a_{3}+3 x b_{1}-5 x b_{2}-4 x b_{3}-3 y a_{1}+5 y a_{2}+4 y a_{3}-4 y b_{2}-2 y b_{3}+2 a_{2}+a_{3}+3 b_{1}-4 b_{2}-2 b_{3}}{\left (2 x +y +2\right )^{2}} = 0
\]
Setting the numerator to zero gives
\begin{equation}
\tag{6E} -2 x^{2} a_{2}-x^{2} a_{3}+x^{2} b_{2}+2 x^{2} b_{3}-2 x y a_{2}-4 x y a_{3}+4 x y b_{2}+2 x y b_{3}-2 y^{2} a_{2}-y^{2} a_{3}+y^{2} b_{2}+2 y^{2} b_{3}-4 x a_{2}-2 x a_{3}-3 x b_{1}+5 x b_{2}+4 x b_{3}+3 y a_{1}-5 y a_{2}-4 y a_{3}+4 y b_{2}+2 y b_{3}-2 a_{2}-a_{3}-3 b_{1}+4 b_{2}+2 b_{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} -2 a_{2} v_{1}^{2}-2 a_{2} v_{1} v_{2}-2 a_{2} v_{2}^{2}-a_{3} v_{1}^{2}-4 a_{3} v_{1} v_{2}-a_{3} v_{2}^{2}+b_{2} v_{1}^{2}+4 b_{2} v_{1} v_{2}+b_{2} v_{2}^{2}+2 b_{3} v_{1}^{2}+2 b_{3} v_{1} v_{2}+2 b_{3} v_{2}^{2}+3 a_{1} v_{2}-4 a_{2} v_{1}-5 a_{2} v_{2}-2 a_{3} v_{1}-4 a_{3} v_{2}-3 b_{1} v_{1}+5 b_{2} v_{1}+4 b_{2} v_{2}+4 b_{3} v_{1}+2 b_{3} v_{2}-2 a_{2}-a_{3}-3 b_{1}+4 b_{2}+2 b_{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} \left (-2 a_{2}-a_{3}+b_{2}+2 b_{3}\right ) v_{1}^{2}+\left (-2 a_{2}-4 a_{3}+4 b_{2}+2 b_{3}\right ) v_{1} v_{2}+\left (-4 a_{2}-2 a_{3}-3 b_{1}+5 b_{2}+4 b_{3}\right ) v_{1}+\left (-2 a_{2}-a_{3}+b_{2}+2 b_{3}\right ) v_{2}^{2}+\left (3 a_{1}-5 a_{2}-4 a_{3}+4 b_{2}+2 b_{3}\right ) v_{2}-2 a_{2}-a_{3}-3 b_{1}+4 b_{2}+2 b_{3} = 0
\end{equation}
Setting each coefficients in (8E) to zero gives the following equations to solve
\begin{align*} -2 a_{2}-4 a_{3}+4 b_{2}+2 b_{3}&=0\\ -2 a_{2}-a_{3}+b_{2}+2 b_{3}&=0\\ 3 a_{1}-5 a_{2}-4 a_{3}+4 b_{2}+2 b_{3}&=0\\ -4 a_{2}-2 a_{3}-3 b_{1}+5 b_{2}+4 b_{3}&=0\\ -2 a_{2}-a_{3}-3 b_{1}+4 b_{2}+2 b_{3}&=0 \end{align*}
Solving the above equations for the unknowns gives
\begin{align*} a_{1}&=b_{3}\\ a_{2}&=b_{3}\\ a_{3}&=b_{2}\\ b_{1}&=b_{2}\\ b_{2}&=b_{2}\\ 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 &= y \\
\eta &= x +1 \\
\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 \\ &= x +1 - \left (\frac {2 y +x +1}{2 x +y +2}\right ) \left (y\right ) \\ &= \frac {2 x^{2}-2 y^{2}+4 x +2}{2 x +y +2}\\ \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 {2 x^{2}-2 y^{2}+4 x +2}{2 x +y +2}}} dy \end{align*}
Which results in
\begin{align*} S&= \frac {\ln \left (x +y +1\right )}{4}-\frac {3 \ln \left (-1-x +y \right )}{4} \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 {2 y +x +1}{2 x +y +2} \end{align*}
Evaluating all the partial derivatives gives
\begin{align*} R_{x} &= 1\\ R_{y} &= 0\\ S_{x} &= \frac {-2 y -x -1}{2 \left (x +y +1\right ) \left (x -y +1\right )}\\ S_{y} &= \frac {2 x +y +2}{2 \left (x +y +1\right ) \left (x -y +1\right )} \end{align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.
\begin{align*} \frac {dS}{dR} &= 0\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} &= 0 \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 {0\, dR} + c_2 \\ S \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 {\ln \left (x +y \left (x \right )+1\right )}{4}-\frac {3 \ln \left (y \left (x \right )-x -1\right )}{4} = 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 \((R,S)\)
\( \frac {dy}{dx} = \frac {2 y +x +1}{2 x +y +2}\)
\( \frac {d S}{d R} = 0\)
\(\!\begin {aligned} R&= x\\ S&= \frac {\ln \left (x +y +1\right )}{4}-\frac {3 \ln \left (-1-x +y \right )}{4} \end {aligned} \)
Figure 2.12: Slope field \(y^{\prime }\left (x \right ) = \frac {2 y \left (x \right )+x +1}{2 x +y \left (x \right )+2}\) Summary of solutions found
\begin{align*}
\frac {\ln \left (x +y \left (x \right )+1\right )}{4}-\frac {3 \ln \left (y \left (x \right )-x -1\right )}{4} &= c_2 \\
\end{align*}
✓ Time used: 0.096 (sec). Leaf size: 117 ode := diff ( y ( x ), x ) = (x+2*y(x)+1)/(2*x+2+y(x));
dsolve ( ode , y ( x ), singsol=all);
\[
y = -\frac {\left (i \sqrt {3}-1\right ) \left (3 \sqrt {3}\, \sqrt {27 c_1^{2} \left (x +1\right )^{2}-1}-27 \left (x +1\right ) c_1 \right )^{{2}/{3}}-3 i \sqrt {3}-3-6 \left (3 \sqrt {3}\, \sqrt {27 c_1^{2} \left (x +1\right )^{2}-1}-27 c_1 x -27 c_1 \right )^{{1}/{3}} \left (x +1\right ) c_1}{6 \left (3 \sqrt {3}\, \sqrt {27 c_1^{2} \left (x +1\right )^{2}-1}-27 \left (x +1\right ) c_1 \right )^{{1}/{3}} c_1}
\]
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 C
trying homogeneous types:
trying homogeneous D
<- homogeneous successful
<- homogeneous successful
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {x +2 y \left (x \right )+1}{2 x +y \left (x \right )+2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {x +2 y \left (x \right )+1}{2 x +y \left (x \right )+2} \end {array} \]
✓ Time used: 60.123 (sec). Leaf size: 1598 ode = D [ y [ x ], x ]==( x +2* y [ x ]+1)/(2* x + y [ x ]+2); ic ={}; DSolve [{ ode , ic }, y [ x ], x , IncludeSingularSolutions -> True ] Too large to display
✓ Time used: 98.379 (sec). Leaf size: 371 from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq((-x - 2*y(x) - 1)/(2*x + y(x) + 2) + Derivative(y(x), x),0)
ics = {}
dsolve ( ode , func = y ( x ), ics = ics ) \[
\left [ y{\left (x \right )} = \frac {\frac {2 \cdot 3^{\frac {2}{3}} i C_{1}}{3 \sqrt [3]{C_{1} \left (9 x + \sqrt {3} \sqrt {C_{1} + 27 x^{2} + 54 x + 27} + 9\right )}} + \sqrt {3} x - i x + \frac {3^{\frac {5}{6}} \sqrt [3]{C_{1} \left (9 x + \sqrt {3} \sqrt {C_{1} + 27 x^{2} + 54 x + 27} + 9\right )}}{3} + \frac {\sqrt [3]{3} i \sqrt [3]{C_{1} \left (9 x + \sqrt {3} \sqrt {C_{1} + 27 x^{2} + 54 x + 27} + 9\right )}}{3} + \sqrt {3} - i}{\sqrt {3} - i}, \ y{\left (x \right )} = \frac {- \frac {2 \cdot 3^{\frac {2}{3}} i C_{1}}{3 \sqrt [3]{C_{1} \left (9 x + \sqrt {3} \sqrt {C_{1} + 27 x^{2} + 54 x + 27} + 9\right )}} + \sqrt {3} x + i x + \frac {3^{\frac {5}{6}} \sqrt [3]{C_{1} \left (9 x + \sqrt {3} \sqrt {C_{1} + 27 x^{2} + 54 x + 27} + 9\right )}}{3} - \frac {\sqrt [3]{3} i \sqrt [3]{C_{1} \left (9 x + \sqrt {3} \sqrt {C_{1} + 27 x^{2} + 54 x + 27} + 9\right )}}{3} + \sqrt {3} + i}{\sqrt {3} + i}, \ y{\left (x \right )} = \frac {3^{\frac {2}{3}} C_{1}}{3 \sqrt [3]{C_{1} \left (9 x + \sqrt {3} \sqrt {C_{1} + 27 x^{2} + 54 x + 27} + 9\right )}} + x - \frac {\sqrt [3]{3} \sqrt [3]{C_{1} \left (9 x + \sqrt {3} \sqrt {C_{1} + 27 x^{2} + 54 x + 27} + 9\right )}}{3} + 1\right ]
\]