2.26.13 Problem 13

Internal problem ID [13649]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.4. Equations Containing Polynomial Functions of y. subsection 1.4.1-2 Abel equations of the first kind.
Problem number : 13
Date solved : Thursday, January 01, 2026 at 02:14:13 AM
CAS classification : [_rational, _Abel]

2.26.13.1 Solved using first_order_ode_exact

1.453 (sec)

Entering first order ode exact solver

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

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

\begin{align*} M(x,y) &= -a \,x^{4} y^{3}-\left (b \,x^{2}-1\right ) y -c x\\ N(x,y) &= x \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 (-a \,x^{4} y^{3}-\left (b \,x^{2}-1\right ) y -c x\right )\\ &= -3 x^{4} a \,y^{2}-b \,x^{2}+1 \end{align*}

And

\begin{align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (x\right )\\ &= 1 \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 ) \\ &=\frac {1}{x}\left ( \left ( -3 x^{4} a \,y^{2}-b \,x^{2}+1\right ) - \left (1 \right ) \right ) \\ &=-x \left (3 x^{2} a \,y^{2}+b \right ) \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 {1}{a \,x^{4} y^{3}+\left (b \,x^{2}-1\right ) y +c x}\left ( \left ( 1\right ) - \left (-3 x^{4} a \,y^{2}-b \,x^{2}+1 \right ) \right ) \\ &=-\frac {x^{2} \left (3 x^{2} a \,y^{2}+b \right )}{a \,x^{4} y^{3}+b \,x^{2} y +c x -y} \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 (1\right )-\left (-3 x^{4} a \,y^{2}-b \,x^{2}+1\right )} {x\left (-a \,x^{4} y^{3}-\left (b \,x^{2}-1\right ) y -c x\right ) - y\left (x\right )} \\ &= \frac {-3 x^{2} a \,y^{2}-b}{x^{3} a \,y^{3}+b x y +c} \end{align*}

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

\[ R = \frac {-3 a \,t^{2}-b}{a \,t^{3}+b t +c} \]

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 a \,t^{2}-b}{a \,t^{3}+b t +c}\right )\mathop {\mathrm {d}t} } \end{align*}

The result of integrating gives

\begin{align*} \mu &= e^{-\ln \left (a \,t^{3}+b t +c \right ) } \\ &= \frac {1}{a \,t^{3}+b t +c} \end{align*}

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

\[ \mu =\frac {1}{x^{3} a \,y^{3}+b x y +c} \]

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 {1}{x^{3} a \,y^{3}+b x y +c}\left (-a \,x^{4} y^{3}-\left (b \,x^{2}-1\right ) y -c x\right ) \\ &= \frac {-a \,x^{4} y^{3}+\left (-b \,x^{2}+1\right ) y -c x}{x^{3} a \,y^{3}+b x y +c} \end{align*}

And

\begin{align*} \overline {N} &=\mu N \\ &= \frac {1}{x^{3} a \,y^{3}+b x y +c}\left (x\right ) \\ &= \frac {x}{x^{3} a \,y^{3}+b x y +c} \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 {-a \,x^{4} y^{3}+\left (-b \,x^{2}+1\right ) y -c x}{x^{3} a \,y^{3}+b x y +c}\right ) + \left (\frac {x}{x^{3} a \,y^{3}+b x y +c}\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 {-a \,x^{4} y^{3}+\left (-b \,x^{2}+1\right ) y -c x}{x^{3} a \,y^{3}+b x y +c}\mathop {\mathrm {d}x} \\ \tag{3} \phi &= \int \frac {-a \,x^{4} y^{3}+\left (-b \,x^{2}+1\right ) y -c x}{x^{3} a \,y^{3}+b x y +c}d 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{equation} \tag{4} \frac {\partial \phi }{\partial y} = \frac {x}{x^{3} a \,y^{3}+b x y +c}+\frac {d}{d y}f \left (y \right )+f'(y) \end{equation}

But equation (2) says that \(\frac {\partial \phi }{\partial y} = \frac {x}{x^{3} a \,y^{3}+b x y +c}\). Therefore equation (4) becomes

\begin{equation} \tag{5} \frac {x}{x^{3} a \,y^{3}+b x y +c} = \frac {x}{x^{3} a \,y^{3}+b x y +c}+\frac {d}{d y}f \left (y \right )+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 \(\phi \)

\[ \phi = \int \frac {-a \,x^{4} y^{3}+\left (-b \,x^{2}+1\right ) y -c x}{x^{3} a \,y^{3}+b x y +c}d x+ c_1 \]

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

\[ c_1 = \int \frac {-a \,x^{4} y^{3}+\left (-b \,x^{2}+1\right ) y -c x}{x^{3} a \,y^{3}+b x y +c}d x \]

Summary of solutions found

\begin{align*} \int _{}^{x}\frac {-a \,\textit {\_a}^{4} y^{3}+\left (-\textit {\_a}^{2} b +1\right ) y-\textit {\_a} c}{\textit {\_a}^{3} a y^{3}+\textit {\_a} b y+c}d \textit {\_a} &= c_1 \\ \end{align*}

2.26.13.2 ✓ Maple. Time used: 0.004 (sec). Leaf size: 53

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

\[ y = -\frac {\operatorname {RootOf}\left (b \,x^{2}-2 b^{3} \int _{}^{\textit {\_Z}}\frac {1}{a \,\textit {\_a}^{3} c^{2}+\textit {\_a} \,b^{3}-b^{3}}d \textit {\_a} +2 c_1 \right ) c}{x b} \]

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 Chini 
<- Chini successful
 

Maple step by step

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

2.26.13.3 ✓ Mathematica. Time used: 0.166 (sec). Leaf size: 317

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

\[ \text {Solve}\left [\frac {1}{3} a c^2 \text {RootSum}\left [\text {$\#$1}^9 a c^2+3 \text {$\#$1}^6 a c^2+3 \text {$\#$1}^3 a c^2+\text {$\#$1}^3 b^3+a c^2\&,\frac {\text {$\#$1}^6 \log \left (y(x) \sqrt [3]{\frac {a x^3}{c}}-\text {$\#$1}\right )+\text {$\#$1}^4 \sqrt [3]{-\frac {b^3}{a c^2}} \log \left (y(x) \sqrt [3]{\frac {a x^3}{c}}-\text {$\#$1}\right )+2 \text {$\#$1}^3 \log \left (y(x) \sqrt [3]{\frac {a x^3}{c}}-\text {$\#$1}\right )+\text {$\#$1}^2 \left (-\frac {b^3}{a c^2}\right )^{2/3} \log \left (y(x) \sqrt [3]{\frac {a x^3}{c}}-\text {$\#$1}\right )+\text {$\#$1} \sqrt [3]{-\frac {b^3}{a c^2}} \log \left (y(x) \sqrt [3]{\frac {a x^3}{c}}-\text {$\#$1}\right )+\log \left (y(x) \sqrt [3]{\frac {a x^3}{c}}-\text {$\#$1}\right )}{3 \text {$\#$1}^8 a c^2+6 \text {$\#$1}^5 a c^2+3 \text {$\#$1}^2 a c^2+\text {$\#$1}^2 b^3}\&\right ]=\frac {1}{2} c x \sqrt [3]{\frac {a x^3}{c}}+c_1,y(x)\right ] \]

2.26.13.4 ✗ Sympy

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

NotImplementedError : The given ODE -a*x**3*y(x)**3 - b*x*y(x) - c + Derivative(y(x), x) + y(x)/x cannot be solved by the factorable group method