60.1.156 problem 157
Internal
problem
ID
[10170]
Book
:
Differential
Gleichungen,
E.
Kamke,
3rd
ed.
Chelsea
Pub.
NY,
1948
Section
:
Chapter
1,
linear
first
order
Problem
number
:
157
Date
solved
:
Tuesday, January 28, 2025 at 04:26:23 PM
CAS
classification
:
[_rational, _Riccati]
\begin{align*} \left (x^{2}-1\right ) y^{\prime }+a \left (y^{2}-2 y x +1\right )&=0 \end{align*}
✓ Solution by Maple
Time used: 0.017 (sec). Leaf size: 280
dsolve((x^2-1)*diff(y(x),x) + a*(y(x)^2-2*x*y(x)+1)=0,y(x), singsol=all)
\[
y = \frac {2 \left (-\frac {x}{2}-\frac {1}{2}\right )^{2 a} \left (-\frac {\left (-\frac {x}{2}-\frac {1}{2}\right )^{-2 a} \left (x +1\right ) \left (x -1\right )^{2} \operatorname {HeunCPrime}\left (0, 2 a -1, 0, 0, a^{2}-a +\frac {1}{2}, \frac {2}{x +1}\right )}{8}-\operatorname {HeunCPrime}\left (0, -2 a +1, 0, 0, a^{2}-a +\frac {1}{2}, \frac {2}{x +1}\right ) c_{1} \left (x -1\right )^{2}+\left (c_{1} \left (\left (a -\frac {1}{2}\right ) x -\frac {a}{2}+\frac {1}{2}\right ) \left (\frac {x +1}{x -1}\right )^{-a} \operatorname {hypergeom}\left (\left [-a +1, -a +1\right ], \left [-2 a +2\right ], -\frac {2}{x -1}\right )+\frac {a \left (\frac {x +1}{x -1}\right )^{a} \left (-\frac {x}{2}-\frac {1}{2}\right )^{-2 a} \operatorname {hypergeom}\left (\left [a , a\right ], \left [2 a \right ], -\frac {2}{x -1}\right ) \left (x -1\right )}{16}\right ) \left (x +1\right )^{2}\right ) \left (\frac {x +1}{x -1}\right )^{a}}{a \left (x +1\right )^{2} \left (c_{1} \operatorname {hypergeom}\left (\left [-a +1, -a +1\right ], \left [-2 a +2\right ], -\frac {2}{x -1}\right ) \left (-\frac {x}{2}-\frac {1}{2}\right )^{2 a}+\frac {\left (\frac {x +1}{x -1}\right )^{2 a} \operatorname {hypergeom}\left (\left [a , a\right ], \left [2 a \right ], -\frac {2}{x -1}\right ) \left (x -1\right )}{8}\right )}
\]
✓ Solution by Mathematica
Time used: 0.357 (sec). Leaf size: 47
DSolve[(x^2-1)*D[y[x],x] + a*(y[x]^2-2*x*y[x]+1)==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*}
y(x)\to \frac {\operatorname {LegendreQ}(a,x)+c_1 \operatorname {LegendreP}(a,x)}{\operatorname {LegendreQ}(a-1,x)+c_1 \operatorname {LegendreP}(a-1,x)} \\
y(x)\to \frac {\operatorname {LegendreP}(a,x)}{\operatorname {LegendreP}(a-1,x)} \\
\end{align*}