2.26.6 Problem 6

Internal problem ID [13642]
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 : 6
Date solved : Thursday, January 01, 2026 at 02:13:21 AM
CAS classification : [_Abel]

2.26.6.1 Solved using first_order_ode_abel_first_kind

4.640 (sec)

Entering first order ode abel first kind solver

\begin{align*} y^{\prime }&=a y^{3}+3 a b x y^{2}-b -2 a \,b^{3} x^{3} \\ \end{align*}

This is Abel first kind ODE, it has the form

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

Comparing the above to given ODE which is

\begin{align*}y^{\prime }&=a y^{3}+3 a b x y^{2}-b -2 a \,b^{3} x^{3}\tag {1} \end{align*}

Therefore

\begin{align*} f_0 &= -2 a \,b^{3} x^{3}-b\\ f_1 &= 0\\ f_2 &= 3 a b x\\ f_3 &= a \end{align*}

Hence

\begin{align*} f'_{0} &= -6 a \,b^{3} x^{2}\\ f'_{3} &= 0 \end{align*}

Since \(f_2(x)=3 a b x\) is not zero, then the followingtransformation is used to remove \(f_2\). Let \(y = u(x) - \frac {f_2}{3 f_3}\) or

\begin{align*} y &= u(x) - \left ( \frac {3 a b x}{3 a} \right ) \\ &= u \left (x \right )-b x \end{align*}

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

\begin{align*} u^{\prime }\left (x \right ) = -3 u \left (x \right ) a \,b^{2} x^{2}+u \left (x \right )^{3} a\tag {2} \end{align*}

The above ODE (2) can now be solved.

Entering first order ode bernoulli solver In canonical form, the ODE is

\begin{align*} u' &= F(x,u)\\ &= \left (-3 u a \,b^{2} x^{2}+u^{3} a \right )_{1} \end{align*}

This is a Bernoulli ODE.

\[ u' = \left (-3 a \,b^{2} x^{2}\right ) u \left (x \right ) + \left (a\right )u^{3} \tag {1} \]

The standard Bernoulli ODE has the form

\[ u' = f_0(x)u+f_1(x)u^n \tag {2} \]

Comparing this to (1) shows that

\begin{align*} f_0 &=-3 a \,b^{2} x^{2}\\ f_1 &=a \end{align*}

The first step is to divide the above equation by \(u^n \) which gives

\[ \frac {u'}{u^n} = f_0(x) u^{1-n} +f_1(x) \tag {3} \]

The next step is use the substitution \(v = u^{1-n}\) in equation (3) which generates a new ODE in \(v \left (x \right )\) which will be linear and can be easily solved using an integrating factor. Backsubstitution then gives the solution \(u(x)\) which is what we want.

This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows that

\begin{align*} f_0(x)&=-3 a \,b^{2} x^{2}\\ f_1(x)&=a\\ n &=3 \end{align*}

Dividing both sides of ODE (1) by \(u^n=u^{3}\) gives

\begin{align*} u'\frac {1}{u^{3}} &= -\frac {3 a \,b^{2} x^{2}}{u^{2}} +a \tag {4} \end{align*}

Let

\begin{align*} v &= u^{1-n} \\ &= \frac {1}{u^{2}} \tag {5} \end{align*}

Taking derivative of equation (5) w.r.t \(x\) gives

\begin{align*} v' &= -\frac {2}{u^{3}}u' \tag {6} \end{align*}

Substituting equations (5) and (6) into equation (4) gives

\begin{align*} -\frac {v^{\prime }\left (x \right )}{2}&= -3 a \,b^{2} x^{2} v \left (x \right )+a\\ v' &= 6 a \,b^{2} x^{2} v -2 a \tag {7} \end{align*}

The above now is a linear ODE in \(v \left (x \right )\) which is now solved.

In canonical form a linear first order is

\begin{align*} v^{\prime }\left (x \right ) + q(x)v \left (x \right ) &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=-6 a \,b^{2} x^{2}\\ p(x) &=-2 a \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -6 a \,b^{2} x^{2}d x}\\ &= {\mathrm e}^{-2 a \,b^{2} x^{3}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu v\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu v\right ) &= \left (\mu \right ) \left (-2 a\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (v \,{\mathrm e}^{-2 a \,b^{2} x^{3}}\right ) &= \left ({\mathrm e}^{-2 a \,b^{2} x^{3}}\right ) \left (-2 a\right ) \\ \mathrm {d} \left (v \,{\mathrm e}^{-2 a \,b^{2} x^{3}}\right ) &= \left (-2 a \,{\mathrm e}^{-2 a \,b^{2} x^{3}}\right )\, \mathrm {d} x \\ \end{align*}

Integrating gives

\begin{align*} v \,{\mathrm e}^{-2 a \,b^{2} x^{3}}&= \int {-2 a \,{\mathrm e}^{-2 a \,b^{2} x^{3}} \,dx} \\ &=\int -2 a \,{\mathrm e}^{-2 a \,b^{2} x^{3}}d x + c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{-2 a \,b^{2} x^{3}}\) gives the final solution

\[ v \left (x \right ) = {\mathrm e}^{2 a \,b^{2} x^{3}} \left (\int -2 a \,{\mathrm e}^{-2 a \,b^{2} x^{3}}d x +c_1 \right ) \]

The substitution \(v = u^{1-n}\) is now used to convert the above solution back to \(u \left (x \right )\) which results in

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

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

\begin{align*} u \left (x \right ) &= \frac {\sqrt {-\left (2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1 \right ) {\mathrm e}^{-2 a \,b^{2} x^{3}}}}{2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1} \\ u \left (x \right ) &= -\frac {\sqrt {-\left (2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1 \right ) {\mathrm e}^{-2 a \,b^{2} x^{3}}}}{2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1} \\ \end{align*}

Now we transform the solution \(u \left (x \right ) = \frac {\sqrt {-\left (2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1 \right ) {\mathrm e}^{-2 a \,b^{2} x^{3}}}}{2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1}\) to \(y\) using \(y = u(x) - \frac {f_2}{3 f_3}\), which gives

\begin{align*} y &= \frac {\sqrt {-\left (2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1 \right ) {\mathrm e}^{-2 a \,b^{2} x^{3}}}}{2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1} - \left (b x\right ) \\ &= \frac {\sqrt {-\left (2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1 \right ) {\mathrm e}^{-2 a \,b^{2} x^{3}}}}{2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1}-b x\\ &= \frac {-2 \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x a b x +c_1 b x +\sqrt {\left (-2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x +c_1 \right ) {\mathrm e}^{-2 a \,b^{2} x^{3}}}}{2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1} \end{align*}

Now we transform the solution \(u \left (x \right ) = -\frac {\sqrt {-\left (2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1 \right ) {\mathrm e}^{-2 a \,b^{2} x^{3}}}}{2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1}\) to \(y\) using \(y = u(x) - \frac {f_2}{3 f_3}\), which gives

\begin{align*} y &= -\frac {\sqrt {-\left (2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1 \right ) {\mathrm e}^{-2 a \,b^{2} x^{3}}}}{2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1} - \left (b x\right ) \\ &= -\frac {\sqrt {-\left (2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1 \right ) {\mathrm e}^{-2 a \,b^{2} x^{3}}}}{2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1}-b x\\ &= \frac {-2 \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x a b x +c_1 b x -\sqrt {\left (-2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x +c_1 \right ) {\mathrm e}^{-2 a \,b^{2} x^{3}}}}{2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1} \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {-2 \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x a b x +c_1 b x -\sqrt {\left (-2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x +c_1 \right ) {\mathrm e}^{-2 a \,b^{2} x^{3}}}}{2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1} \\ y &= \frac {-2 \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x a b x +c_1 b x +\sqrt {\left (-2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x +c_1 \right ) {\mathrm e}^{-2 a \,b^{2} x^{3}}}}{2 a \int {\mathrm e}^{-2 a \,b^{2} x^{3}}d x -c_1} \\ \end{align*}

2.26.6.2 ✓ Maple. Time used: 0.001 (sec). Leaf size: 208

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

\begin{align*} y &= -b x \\ y &= -\frac {2 \,{\mathrm e}^{-a \,x^{3} b^{2}}}{\sqrt {\frac {-3 \,{\mathrm e}^{-a \,x^{3} b^{2}} a x 2^{{5}/{6}} \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, 2 a \,x^{3} b^{2}\right )+4 c_1 \left (a \,x^{3} b^{2}\right )^{{1}/{6}}-8 \,{\mathrm e}^{-2 a \,x^{3} b^{2}} \left (a \,x^{3} b^{2}\right )^{{1}/{6}} a x}{\left (a \,x^{3} b^{2}\right )^{{1}/{6}}}}}-b x \\ y &= \frac {2 \,{\mathrm e}^{-a \,x^{3} b^{2}}}{\sqrt {\frac {-3 \,{\mathrm e}^{-a \,x^{3} b^{2}} a x 2^{{5}/{6}} \operatorname {WhittakerM}\left (\frac {1}{6}, \frac {2}{3}, 2 a \,x^{3} b^{2}\right )+4 c_1 \left (a \,x^{3} b^{2}\right )^{{1}/{6}}-8 \,{\mathrm e}^{-2 a \,x^{3} b^{2}} \left (a \,x^{3} b^{2}\right )^{{1}/{6}} a x}{\left (a \,x^{3} b^{2}\right )^{{1}/{6}}}}}-b x \\ \end{align*}

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 
differential order: 1; looking for linear symmetries 
trying exact 
trying Abel 
<- Abel successful
 

Maple step by step

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

2.26.6.3 ✓ Mathematica. Time used: 4.472 (sec). Leaf size: 138

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

\begin{align*} y(x)&\to -b x-\frac {e^{-a b^2 x^3}}{\sqrt {\frac {2^{2/3} a x \Gamma \left (\frac {1}{3},2 a b^2 x^3\right )}{3 \sqrt [3]{a b^2 x^3}}+c_1}}\\ y(x)&\to -b x+\frac {e^{-a b^2 x^3}}{\sqrt {\frac {2^{2/3} a x \Gamma \left (\frac {1}{3},2 a b^2 x^3\right )}{3 \sqrt [3]{a b^2 x^3}}+c_1}}\\ y(x)&\to -b x \end{align*}

2.26.6.4 ✗ Sympy

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

RecursionError : maximum recursion depth exceeded