2.28.46 Problem 56

Internal problem ID [13717]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.2-2
Problem number : 56
Date solved : Thursday, January 01, 2026 at 02:22:54 AM
CAS classification : [[_2nd_order, _with_linear_symmetries], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]

2.28.46.1 second order linear exact ode

0.878 (sec)

\begin{align*} y^{\prime \prime }+\left (x^{n} a +b \,x^{m}\right ) y^{\prime }+\left (a n \,x^{n -1}+b m \,x^{m -1}\right ) y&=0 \\ \end{align*}

Entering second order linear exact ode solverAn ode of the form

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

is exact if

\begin{align*} p''(x) - q'(x) + r(x) &= 0 \tag {1} \end{align*}

For the given ode we have

\begin{align*} p(x) &= 1\\ q(x) &= x^{n} a +b \,x^{m}\\ r(x) &= \frac {a n \,x^{n}+b \,x^{m} m}{x}\\ s(x) &= 0 \end{align*}

Hence

\begin{align*} p''(x) &= 0\\ q'(x) &= \frac {a n \,x^{n}}{x}+\frac {b \,x^{m} m}{x} \end{align*}

Therefore (1) becomes

\begin{align*} 0- \left (\frac {a n \,x^{n}}{x}+\frac {b \,x^{m} m}{x}\right ) + \left (\frac {a n \,x^{n}+b \,x^{m} m}{x}\right )&=0 \end{align*}

Hence the ode is exact. Since we now know the ode is exact, it can be written as

\begin{align*} \left (p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y\right )' &= s(x) \end{align*}

Integrating gives

\begin{align*} p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y&=\int {s \left (x \right )\, dx} \end{align*}

Substituting the above values for \(p,q,r,s\) gives

\begin{align*} y^{\prime }+\left (x^{n} a +b \,x^{m}\right ) y&=c_1 \end{align*}

We now have a first order ode to solve which is

\begin{align*} y^{\prime }+\left (x^{n} a +b \,x^{m}\right ) y = c_1 \end{align*}

Entering first order ode linear solverIn canonical form a linear first order is

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

Comparing the above to the given ode shows that

\begin{align*} q(x) &=x^{n} a +b \,x^{m}\\ p(x) &=c_1 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \left (x^{n} a +b \,x^{m}\right )d x}\\ &= {\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (c_1\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (y \,{\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}\right ) &= \left ({\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}\right ) \left (c_1\right ) \\ \mathrm {d} \left (y \,{\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}\right ) &= \left (c_1 \,{\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}\right )\, \mathrm {d} x \\ \end{align*}

Integrating gives

\begin{align*} y \,{\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}&= \int {c_1 \,{\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}} \,dx} \\ &=\int c_1 \,{\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}d x + c_2 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}\) gives the final solution

\[ y = {\mathrm e}^{-\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}} \left (\int c_1 \,{\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}d x +c_2 \right ) \]

Summary of solutions found

\begin{align*} y &= {\mathrm e}^{-\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}} \left (\int c_1 \,{\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}d x +c_2 \right ) \\ \end{align*}

2.28.46.2 second order integrable as is

0.546 (sec)

\begin{align*} y^{\prime \prime }+\left (x^{n} a +b \,x^{m}\right ) y^{\prime }+\left (a n \,x^{n -1}+b m \,x^{m -1}\right ) y&=0 \\ \end{align*}

Entering second order integrable as is solverIntegrating both sides of the ODE w.r.t \(x\) gives

\begin{align*} \int \left (y^{\prime \prime }+\left (x^{n} a +b \,x^{m}\right ) y^{\prime }+\frac {\left (a n \,x^{n}+b \,x^{m} m \right ) y}{x}\right )d x &= 0 \\ \frac {\left (x^{n} a x +x^{m} b x \right ) y}{x}+y^{\prime } = c_1 \end{align*}

Which is now solved for \(y\). Entering first order ode linear solverIn canonical form a linear first order is

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

Comparing the above to the given ode shows that

\begin{align*} q(x) &=x^{n} a +b \,x^{m}\\ p(x) &=c_1 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \left (x^{n} a +b \,x^{m}\right )d x}\\ &= {\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (c_1\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (y \,{\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}\right ) &= \left ({\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}\right ) \left (c_1\right ) \\ \mathrm {d} \left (y \,{\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}\right ) &= \left (c_1 \,{\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}\right )\, \mathrm {d} x \\ \end{align*}

Integrating gives

\begin{align*} y \,{\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}&= \int {c_1 \,{\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}} \,dx} \\ &=\int c_1 \,{\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}d x + c_2 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}\) gives the final solution

\[ y = {\mathrm e}^{-\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}} \left (\int c_1 \,{\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}d x +c_2 \right ) \]

Summary of solutions found

\begin{align*} y &= {\mathrm e}^{-\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}} \left (\int c_1 \,{\mathrm e}^{\frac {x \left (a \left (1+m \right ) x^{n}+b \left (n +1\right ) x^{m}\right )}{\left (n +1\right ) \left (1+m \right )}}d x +c_2 \right ) \\ \end{align*}

2.28.46.3 ✓ Maple. Time used: 0.009 (sec). Leaf size: 72

ode:=diff(diff(y(x),x),x)+(a*x^n+b*x^m)*diff(y(x),x)+(a*n*x^(n-1)+b*m*x^(m-1))*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = \left (c_1 \int {\mathrm e}^{\frac {x \left (a \,x^{n} \left (m +1\right )+b \,x^{m} \left (n +1\right )\right )}{\left (n +1\right ) \left (m +1\right )}}d x +c_2 \right ) {\mathrm e}^{-\frac {x \left (a \,x^{n} \left (m +1\right )+b \,x^{m} \left (n +1\right )\right )}{\left (n +1\right ) \left (m +1\right )}} \]

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
   One independent solution has integrals. Trying a hypergeometric solution fre\ 
e of integrals... 
   -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
No hypergeometric solution was found. 
<- linear_1 successful
 

2.28.46.4 ✓ Mathematica. Time used: 60.052 (sec). Leaf size: 74

ode=D[y[x],{x,2}]+(a*x^n+b*x^m)*D[y[x],x]+(a*n*x^(n-1)+b*m*x^(m-1))*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to e^{x \left (-\frac {a x^n}{n+1}-\frac {b x^m}{m+1}\right )} \left (\int _1^x\exp \left (K[1] \left (\frac {b K[1]^m}{m+1}+\frac {a K[1]^n}{n+1}\right )\right ) c_1dK[1]+c_2\right ) \end{align*}

2.28.46.5 ✗ Sympy

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

TypeError : Add object cannot be interpreted as an integer