54.3.350 problem 1367

Internal problem ID [12645]
Book : Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section : Chapter 2, linear second order
Problem number : 1367
Date solved : Friday, October 03, 2025 at 03:45:31 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

\begin{align*} y^{\prime \prime }&=-\frac {2 x y^{\prime }}{x^{2}+1}-\frac {\left (a^{2} \left (x^{2}+1\right )^{2}-n \left (n +1\right ) \left (x^{2}+1\right )+m^{2}\right ) y}{\left (x^{2}+1\right )^{2}} \end{align*}

✓ Maple. Time used: 0.098 (sec). Leaf size: 88

ode:=diff(diff(y(x),x),x) = -2/(x^2+1)*x*diff(y(x),x)-(a^2*(x^2+1)^2-n*(n+1)*(x^2+1)+m^2)/(x^2+1)^2*y(x); 
dsolve(ode,y(x), singsol=all);
 

\[ y = \left (\operatorname {HeunC}\left (0, \frac {1}{2}, m , -\frac {a^{2}}{4}, \frac {1}{4}+\frac {1}{4} a^{2}+\frac {1}{4} m^{2}-\frac {1}{4} n^{2}-\frac {1}{4} n , -x^{2}\right ) c_2 x +\operatorname {HeunC}\left (0, -\frac {1}{2}, m , -\frac {a^{2}}{4}, \frac {1}{4}+\frac {1}{4} a^{2}+\frac {1}{4} m^{2}-\frac {1}{4} n^{2}-\frac {1}{4} n , -x^{2}\right ) c_1 \right ) \left (x^{2}+1\right )^{\frac {m}{2}} \]

✓ Mathematica. Time used: 0.229 (sec). Leaf size: 140

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

\begin{align*} y(x)&\to \left (x^2+1\right )^{\frac {\sqrt {m^2}}{2}} \left (c_2 x \text {HeunC}\left [\frac {1}{4} \left (-a^2-m^2-3 \sqrt {m^2}+n^2+n-2\right ),-\frac {a^2}{4},\frac {3}{2},\sqrt {m^2}+1,0,-x^2\right ]+c_1 \text {HeunC}\left [\frac {1}{4} \left (-a^2-m^2-\sqrt {m^2}+n^2+n\right ),-\frac {a^2}{4},\frac {1}{2},\sqrt {m^2}+1,0,-x^2\right ]\right ) \end{align*}

✗ Sympy

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

False