2.35.1 Problem 1

Internal problem ID [13925]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Differential Equations. section 2.1.3-1. Equations with exponential functions
Problem number : 1
Date solved : Sunday, January 18, 2026 at 09:34:45 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

2.35.1.1 second order bessel ode form A

0.048 (sec)

\begin{align*} y^{\prime \prime }+a \,{\mathrm e}^{\lambda x} y&=0 \\ \end{align*}

Entering second order bessel ode form A solverWriting the ode as

\begin{align*} y^{\prime \prime }+a \,{\mathrm e}^{\lambda x} y = 0\tag {1} \end{align*}

An ode of the form

\begin{equation} ay^{\prime \prime }+by^{\prime }+(ce^{rx}+m)y=0\tag {1}\end{equation}

can be transformed to Bessel ode using the transformation

\begin{align*} x & =\ln \left ( t\right ) \\ e^{x} & =t \end{align*}

Where \(a,b,c,m\) are not functions of \(x\) and where \(b\) and \(m\) are allowed to be be zero. Using this transformation gives

\begin{align} \frac {dy}{dx} & =\frac {dy}{dt}\frac {dt}{dx}\nonumber \\ & =\frac {dy}{dt}e^{x}\nonumber \\ & =t\frac {dy}{dt}\tag {2}\end{align}

And

\begin{align} \frac {d^{2}y}{dx^{2}} & =\frac {d}{dx}\left ( \frac {dy}{dx}\right ) \nonumber \\ & =\frac {d}{dx}\left ( t\frac {dy}{dt}\right ) \nonumber \\ & =\frac {d}{dt}\frac {dt}{dx}\left ( t\frac {dy}{dt}\right ) \nonumber \\ & =\frac {dt}{dx}\frac {d}{dt}\left ( t\frac {dy}{dt}\right ) \nonumber \\ & =t\frac {d}{dt}\left ( t\frac {dy}{dt}\right ) \nonumber \\ & =t\left ( \frac {dy}{dt}+t\frac {d^{2}y}{dt^{2}}\right ) \tag {3}\end{align}

Substituting (2,3) into (1) gives

\begin{align} at\left ( \frac {dy}{dt}+t\frac {d^{2}y}{dt^{2}}\right ) +bt\frac {dy}{dt}+(ce^{rx}+m)y & =0\nonumber \\ \left ( aty^{\prime }+at^{2}y^{\prime \prime }\right ) +bty^{\prime }+(ct^{r}+m)y & =0\nonumber \\ at^{2}y^{\prime \prime }+\left ( b+a\right ) ty^{\prime }+(ct^{r}+m)y & =0\nonumber \\ t^{2}y^{\prime \prime }+\frac {b+a}{a}ty^{\prime }+\left ( \frac {c}{a}t^{r}+\frac {m}{a}\right ) y & =0\tag {4}\end{align}

Which is Bessel ODE. Comparing the above to the general known Bowman form of Bessel ode which is

\begin{equation} t^{2}y^{\prime \prime }+\left ( 1-2\alpha \right ) ty^{\prime }+\left ( \beta ^{2}\gamma ^{2}t^{2\gamma }-\left ( n^{2}\gamma ^{2}-\alpha ^{2}\right ) \right ) y=0\tag {C}\end{equation}

And now comparing (4) and (C) shows that

\begin{align} \left ( 1-2\alpha \right ) & =\frac {b+a}{a}\tag {5}\\ \beta ^{2}\gamma ^{2} & =\frac {c}{a}\tag {6}\\ 2\gamma & =r\tag {7}\\ \left ( n^{2}\gamma ^{2}-\alpha ^{2}\right ) & =-\frac {m}{a}\tag {8}\end{align}

(5) gives \(\alpha =\frac {1}{2}-\frac {b+a}{2a}\). (7) gives \(\gamma =\frac {r}{2}\). (8) now becomes \(\left ( n^{2}\left ( \frac {r}{2}\right ) ^{2}-\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}\right ) =-\frac {m}{a}\) or \(n^{2}=\frac {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}{\left ( \frac {r}{2}\right ) ^{2}}\). Hence \(n=\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}\ \)by taking the positive root. And finally (6) gives \(\beta ^{2}=\frac {c}{a\gamma ^{2}}\) or \(\beta =\sqrt {\frac {c}{a}}\frac {1}{\gamma }=\sqrt {\frac {c}{a}}\frac {2}{r}\) (also taking the positive root). Hence

\begin{align*} \alpha & =\frac {1}{2}-\frac {b+a}{2a}\\ n & =\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}\\ \beta & =\sqrt {\frac {c}{a}}\frac {2}{r}\\ \gamma & =\frac {r}{2}\end{align*}

But the solution to (C) which is general form of Bessel ode is known and given by

\[ y\left ( t\right ) =t^{\alpha }\left ( c_{1}J_{n}\left ( \beta t^{\gamma }\right ) +c_{2}Y_{n}\left ( \beta t^{\gamma }\right ) \right ) \]

Substituting the above values found into this solution gives

\[ y\left ( t\right ) =t^{\frac {1}{2}-\frac {b+a}{2a}}\left ( c_{1}J_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}t^{\frac {r}{2}}\right ) +c_{2}Y_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}t^{\frac {r}{2}}\right ) \right ) \]

Since \(e^{x}=t\) then the above becomes

\begin{align} y\left ( x\right ) & =e^{x\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) }\left ( c_{1}J_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) +c_{2}Y_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) \right ) \nonumber \\ & =e^{x\left ( \frac {-b}{2a}\right ) }\left ( c_{1}J_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {-b}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) +c_{2}Y_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {-b}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) \right ) \nonumber \\ & =e^{x\left ( \frac {-b}{2a}\right ) }\left ( c_{1}J_{\frac {2}{r}\sqrt {-\frac {m}{a}+\frac {b^{2}}{4a^{2}}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) +c_{2}Y_{\frac {2}{r}\sqrt {-\frac {m}{a}+\frac {b^{2}}{4a^{2}}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) \right ) \nonumber \\ & =e^{x\left ( \frac {-b}{2a}\right ) }\left ( c_{1}J_{\frac {2}{r}\sqrt {-\frac {4ma+b^{2}}{4a^{2}}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) +c_{2}Y_{\frac {2}{r}\sqrt {-\frac {4ma+b^{2}}{4a^{2}}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) \right ) \nonumber \\ & =e^{x\left ( \frac {-b}{2a}\right ) }\left ( c_{1}J_{\frac {1}{ra}\sqrt {-4ma+b^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) +c_{2}Y_{\frac {1}{ra}\sqrt {-4ma+b^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) \right ) \tag {9}\end{align}

Equation (9) above is the solution to \(ay^{\prime \prime }+by^{\prime }+(ce^{rx}+m)y=0\). Therefore we just need now to compare this form to the ode given and use (9) to obtain the final solution.

Comparing form (1) to the ode we are solving shows that

\begin{align*} a &= 1\\ b &= 0\\ c &= a\\ r &= \lambda \\ m &= 0 \end{align*}

Substituting these in (9) gives the solution as

\begin{align*} y&=c_1 \operatorname {BesselJ}\left (0, \frac {2 \sqrt {a}\, {\mathrm e}^{\frac {\lambda x}{2}}}{\lambda }\right )+c_2 \operatorname {BesselY}\left (0, \frac {2 \sqrt {a}\, {\mathrm e}^{\frac {\lambda x}{2}}}{\lambda }\right ) \end{align*}

Summary of solutions found

\begin{align*} y &= c_1 \operatorname {BesselJ}\left (0, \frac {2 \sqrt {a}\, {\mathrm e}^{\frac {\lambda x}{2}}}{\lambda }\right )+c_2 \operatorname {BesselY}\left (0, \frac {2 \sqrt {a}\, {\mathrm e}^{\frac {\lambda x}{2}}}{\lambda }\right ) \\ \end{align*}

2.35.1.2 ✓ Maple. Time used: 0.014 (sec). Leaf size: 39

ode:=diff(diff(y(x),x),x)+a*exp(lambda*x)*y(x) = 0; 
dsolve(ode,y(x), singsol=all);
 

\[ y = c_1 \operatorname {BesselJ}\left (0, \frac {2 \sqrt {a}\, {\mathrm e}^{\frac {\lambda x}{2}}}{\lambda }\right )+c_2 \operatorname {BesselY}\left (0, \frac {2 \sqrt {a}\, {\mathrm e}^{\frac {\lambda x}{2}}}{\lambda }\right ) \]

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power \ 
@ Moebius 
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x),\ 
 dx)) * 2F1([a1, a2], [b1], f) 
-> Trying changes of variables to rationalize or make the ODE simpler 
   trying a quadrature 
   checking if the LODE has constant coefficients 
   checking if the LODE is of Euler type 
   trying a symmetry of the form [xi=0, eta=F(x)] 
   checking if the LODE is missing y 
   -> Trying a Liouvillian solution using Kovacics algorithm 
   <- No Liouvillian solutions exists 
   -> Trying a solution in terms of special functions: 
      -> Bessel 
      <- Bessel successful 
   <- special function solution successful 
   Change of variables used: 
      [x = ln(t)/lambda] 
   Linear ODE actually solved: 
      a*u(t)+lambda^2*diff(u(t),t)+lambda^2*t*diff(diff(u(t),t),t) = 0 
<- change of variables successful
 

2.35.1.3 ✓ Mathematica. Time used: 0.025 (sec). Leaf size: 55

ode=D[y[x],{x,2}]+a*Exp[\[Lambda]*x]*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to c_1 \operatorname {BesselJ}\left (0,\frac {2 \sqrt {a} \sqrt {e^{x \lambda }}}{\lambda }\right )+2 c_2 \operatorname {BesselY}\left (0,\frac {2 \sqrt {a} \sqrt {e^{x \lambda }}}{\lambda }\right ) \end{align*}

2.35.1.4 ✗ Sympy

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

False
 

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0
 

classify_ode(ode,func=y(x)) 
 
('2nd_power_series_ordinary',)