54.3.355 problem 1372

Internal problem ID [12650]
Book : Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section : Chapter 2, linear second order
Problem number : 1372
Date solved : Friday, October 03, 2025 at 03:45:34 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

\begin{align*} y^{\prime \prime }&=-\frac {2 x y^{\prime }}{x^{2}-1}-\frac {\left (\left (x^{2}-1\right ) \left (a \,x^{2}+b x +c \right )-k^{2}\right ) y}{\left (x^{2}-1\right )^{2}} \end{align*}

✓ Maple. Time used: 0.139 (sec). Leaf size: 101

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

\[ y = {\mathrm e}^{\sqrt {-a}\, x} \left (c_2 \operatorname {HeunC}\left (4 \sqrt {-a}, -k , k , 2 b , \frac {k^{2}}{2}+a -b +c , \frac {x}{2}+\frac {1}{2}\right ) \left (x +1\right )^{-\frac {k}{2}} \left (x -1\right )^{\frac {k}{2}}+c_1 \operatorname {HeunC}\left (4 \sqrt {-a}, k , k , 2 b , \frac {k^{2}}{2}+a -b +c , \frac {x}{2}+\frac {1}{2}\right ) \left (x^{2}-1\right )^{\frac {k}{2}}\right ) \]

✓ Mathematica. Time used: 0.394 (sec). Leaf size: 189

ode=D[y[x],{x,2}] == -(((-k^2 + (-1 + x^2)*(c + b*x + a*x^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 e^{\sqrt {-a} x} (x+1)^{-k/2} \left (c_1 (x+1)^{k/2} \left (x^2-1\right )^{k/2} \text {HeunC}\left [(k+1) \left (2 \sqrt {-a}-k\right )-a+b-c,2 \left (2 \sqrt {-a} (k+1)+b\right ),k+1,k+1,4 \sqrt {-a},\frac {x+1}{2}\right ]+\sqrt {2} c_2 (x-1)^{k/2} \text {HeunC}\left [-2 \sqrt {-a} (k-1)-a+b-c,2 \left (2 \sqrt {-a}+b\right ),1-k,k+1,4 \sqrt {-a},\frac {x+1}{2}\right ]\right ) \end{align*}

✗ Sympy

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

False