2.1.36 Problem 37

Internal problem ID [10022]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 37
Date solved : Monday, December 08, 2025 at 07:05:45 PM
CAS classification : [_rational, _Riccati]

2.1.36.1 Solved using first_order_ode_riccati

1.423 (sec)

Entering first order ode riccati solver

\begin{align*} y^{\prime } x -2 y+b y^{2}&=c \,x^{4} \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= -\frac {-c \,x^{4}+b \,y^{2}-2 y}{x} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = x^{3} c -\frac {b \,y^{2}}{x}+\frac {2 y}{x} \]

With Riccati ODE standard form

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

Shows that \(f_0(x)=x^{3} c\), \(f_1(x)=\frac {2}{x}\) and \(f_2(x)=-\frac {b}{x}\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {b u}{x}} \tag {1} \end{align*}

Using the above substitution in the given ODE results (after some simplification) in a second order ODE to solve for \(u(x)\) which is

\begin{align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end{align*}

But

\begin{align*} f_2' &=\frac {b}{x^{2}}\\ f_1 f_2 &=-\frac {2 b}{x^{2}}\\ f_2^2 f_0 &=b^{2} x c \end{align*}

Substituting the above terms back in equation (2) gives

\[ -\frac {b u^{\prime \prime }\left (x \right )}{x}+\frac {b u^{\prime }\left (x \right )}{x^{2}}+b^{2} x c u \left (x \right ) = 0 \]

Entering second order change of variable on \(x\) method 2 solverIn normal form the ode

\begin{align*} -\frac {b \left (\frac {d^{2}u}{d x^{2}}\right )}{x}+\frac {b \left (\frac {d u}{d x}\right )}{x^{2}}+b^{2} x c u = 0\tag {1} \end{align*}

Becomes

\begin{align*} \frac {d^{2}u}{d x^{2}}+p \left (x \right ) \left (\frac {d u}{d x}\right )+q \left (x \right ) u&=0 \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=-\frac {1}{x}\\ q \left (x \right )&=-b c \,x^{2} \end{align*}

Applying change of variables \(\tau = g \left (x \right )\) to (2) gives

\begin{align*} \frac {d^{2}}{d \tau ^{2}}u \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }u \left (\tau \right )\right )+q_{1} u \left (\tau \right )&=0 \tag {3} \end{align*}

Where \(\tau \) is the new independent variable, and

\begin{align*} p_{1} \left (\tau \right ) &=\frac {\frac {d^{2}}{d x^{2}}\tau \left (x \right )+p \left (x \right ) \left (\frac {d}{d x}\tau \left (x \right )\right )}{\left (\frac {d}{d x}\tau \left (x \right )\right )^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{\left (\frac {d}{d x}\tau \left (x \right )\right )^{2}}\tag {5} \end{align*}

Let \(p_{1} = 0\). Eq (4) simplifies to

\begin{align*} \frac {d^{2}}{d x^{2}}\tau \left (x \right )+p \left (x \right ) \left (\frac {d}{d x}\tau \left (x \right )\right )&=0 \end{align*}

This ode is solved resulting in

\begin{align*} \tau &= \int {\mathrm e}^{-\int p \left (x \right )d x}d x\\ &= \int {\mathrm e}^{-\int -\frac {1}{x}d x}d x\\ &= \int e^{\ln \left (x \right )} \,dx\\ &= \int x d x\\ &= \frac {x^{2}}{2}\tag {6} \end{align*}

Using (6) to evaluate \(q_{1}\) from (5) gives

\begin{align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{\left (\frac {d}{d x}\tau \left (x \right )\right )^{2}}\\ &= \frac {-b c \,x^{2}}{x^{2}}\\ &= -b c\tag {7} \end{align*}

Substituting the above in (3) and noting that now \(p_{1} = 0\) results in

\begin{align*} \frac {d^{2}}{d \tau ^{2}}u \left (\tau \right )+q_{1} u \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}u \left (\tau \right )-b c u \left (\tau \right )&=0 \end{align*}

The above ode is now solved for \(u \left (\tau \right )\).Entering second order linear constant coefficient ode solver

This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

\[ A u''(\tau ) + B u'(\tau ) + C u(\tau ) = 0 \]

Where in the above \(A=1, B=0, C=-b c\). Let the solution be \(u \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives

\[ \lambda ^{2} {\mathrm e}^{\tau \lambda }-b c \,{\mathrm e}^{\tau \lambda } = 0 \tag {1} \]

Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives

\[ -b c +\lambda ^{2} = 0 \tag {2} \]

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]

Substituting \(A=1, B=0, C=-b c\) into the above gives

\begin{align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (-b c\right )}\\ &= \pm \sqrt {b c} \end{align*}

Hence

\begin{align*} \lambda _1 &= + \sqrt {b c} \\ \lambda _2 &= - \sqrt {b c} \\ \end{align*}

Which simplifies to

\begin{align*} \lambda _1 &= \sqrt {b c} \\ \lambda _2 &= -\sqrt {b c} \\ \end{align*}

Since roots are distinct, then the solution is

\begin{align*} u \left (\tau \right ) &= c_1 e^{\lambda _1 \tau } + c_2 e^{\lambda _2 \tau } \\ u \left (\tau \right ) &= c_1 e^{\left (\sqrt {b c}\right )\tau } +c_2 e^{\left (-\sqrt {b c}\right )\tau } \\ \end{align*}

Or

\[ u \left (\tau \right ) =c_1 \,{\mathrm e}^{\sqrt {b c}\, \tau }+c_2 \,{\mathrm e}^{-\sqrt {b c}\, \tau } \]

The above solution is now transformed back to \(u\) using (6) which results in

\[ u = c_1 \,{\mathrm e}^{\frac {\sqrt {b c}\, x^{2}}{2}}+c_2 \,{\mathrm e}^{-\frac {\sqrt {b c}\, x^{2}}{2}} \]

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = c_1 \sqrt {b c}\, x \,{\mathrm e}^{\frac {\sqrt {b c}\, x^{2}}{2}}-c_2 \sqrt {b c}\, x \,{\mathrm e}^{-\frac {\sqrt {b c}\, x^{2}}{2}} \end{equation}

Substituting equations (3,4) into (1) results in

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{-\frac {b u}{x}} \\ y &= \frac {\sqrt {b c}\, x^{2} \left (c_1 \,{\mathrm e}^{\sqrt {b c}\, x^{2}}-c_2 \right )}{b \left (c_1 \,{\mathrm e}^{\sqrt {b c}\, x^{2}}+c_2 \right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = \frac {\left (\sqrt {b c}\, x \,{\mathrm e}^{\frac {\sqrt {b c}\, x^{2}}{2}}-c_3 \sqrt {b c}\, x \,{\mathrm e}^{-\frac {\sqrt {b c}\, x^{2}}{2}}\right ) x}{b \left ({\mathrm e}^{\frac {\sqrt {b c}\, x^{2}}{2}}+c_3 \,{\mathrm e}^{-\frac {\sqrt {b c}\, x^{2}}{2}}\right )} \]

Simplifying the above gives

\begin{align*} y &= \frac {\sqrt {b c}\, x^{2} \left ({\mathrm e}^{\sqrt {b c}\, x^{2}}-c_3 \right )}{b \left ({\mathrm e}^{\sqrt {b c}\, x^{2}}+c_3 \right )} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {\sqrt {b c}\, x^{2} \left ({\mathrm e}^{\sqrt {b c}\, x^{2}}-c_3 \right )}{b \left ({\mathrm e}^{\sqrt {b c}\, x^{2}}+c_3 \right )} \\ \end{align*}

2.1.36.2 Solved using first_order_ode_riccati_by_guessing_particular_solution

0.752 (sec)

Entering first order ode riccati guess solver

\begin{align*} y^{\prime } x -2 y+b y^{2}&=c \,x^{4} \\ \end{align*}

This is a Riccati ODE. Comparing the above ODE to solve with the Riccati standard form

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

Shows that

\begin{align*} f_0(x) & =x^{3} c\\ f_1(x) & =\frac {2}{x}\\ f_2(x) &=-\frac {b}{x} \end{align*}

Using trial and error, the following particular solution was found

\[ y_p = \frac {\sqrt {b c}\, x^{2}}{b} \]

Since a particular solution is known, then the general solution is given by

\begin{align*} y &= y_p + \frac { \phi (x) }{ c_1 - \int { \phi (x) f_2 \,dx} } \end{align*}

Where

\begin{align*} \phi (x) &= e^{ \int 2 f_2 y_p + f_1 \,dx } \end{align*}

Evaluating the above gives the general solution as

\[ y = \frac {\left (-{\mathrm e}^{-\sqrt {b c}\, x^{2}} b -2 \sqrt {b c}\, c_1 \right ) x^{2} c}{b \left ({\mathrm e}^{-\sqrt {b c}\, x^{2}} \sqrt {b c}-2 c_1 c \right )} \]

Summary of solutions found

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

2.1.36.3 ✓ Maple. Time used: 0.004 (sec). Leaf size: 31

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

\[ y = \frac {i \tan \left (-\frac {i x^{2} \sqrt {b}\, \sqrt {c}}{2}+c_1 \right ) x^{2} \sqrt {c}}{\sqrt {b}} \]

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

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

2.1.36.4 ✓ Mathematica. Time used: 0.148 (sec). Leaf size: 153

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

\begin{align*} y(x)&\to \frac {\sqrt {c} x^2 \left (-\cos \left (\frac {1}{2} \sqrt {-b} \sqrt {c} x^2\right )+c_1 \sin \left (\frac {1}{2} \sqrt {-b} \sqrt {c} x^2\right )\right )}{\sqrt {-b} \left (\sin \left (\frac {1}{2} \sqrt {-b} \sqrt {c} x^2\right )+c_1 \cos \left (\frac {1}{2} \sqrt {-b} \sqrt {c} x^2\right )\right )}\\ y(x)&\to \frac {\sqrt {c} x^2 \tan \left (\frac {1}{2} \sqrt {-b} \sqrt {c} x^2\right )}{\sqrt {-b}} \end{align*}

2.1.36.5 ✗ Sympy

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

RecursionError : maximum recursion depth exceeded