29.11.7 problem 298
Internal
problem
ID
[4898]
Book
:
Ordinary
differential
equations
and
their
solutions.
By
George
Moseley
Murphy.
1960
Section
:
Various
11
Problem
number
:
298
Date
solved
:
Tuesday, January 28, 2025 at 02:40:09 PM
CAS
classification
:
[_rational, _Riccati]
\begin{align*} \left (-x^{2}+1\right ) y^{\prime }&=n \left (1-2 y x +y^{2}\right ) \end{align*}
✓ Solution by Maple
Time used: 0.018 (sec). Leaf size: 280
dsolve((-x^2+1)*diff(y(x),x) = n*(1-2*x*y(x)+y(x)^2),y(x), singsol=all)
\[
y \left (x \right ) = -\frac {2 \left (\frac {x +1}{x -1}\right )^{n} \left (-8 \left (x +1\right )^{2} c_{1} \operatorname {hypergeom}\left (\left [1-n , 1-n \right ], \left [-2 n +2\right ], -\frac {2}{x -1}\right ) \left (\left (x -\frac {1}{2}\right ) n +\frac {1}{2}-\frac {x}{2}\right ) \left (\frac {x +1}{x -1}\right )^{-n}+\left (x -1\right ) \left (-\frac {\left (x +1\right )^{2} n \left (\frac {x +1}{x -1}\right )^{n} \left (-\frac {x}{2}-\frac {1}{2}\right )^{-2 n} \operatorname {hypergeom}\left (\left [n , n\right ], \left [2 n \right ], -\frac {2}{x -1}\right )}{2}+\left (x -1\right ) \left (\operatorname {HeunCPrime}\left (0, 2 n -1, 0, 0, n^{2}-n +\frac {1}{2}, \frac {2}{x +1}\right ) \left (x +1\right ) \left (-\frac {x}{2}-\frac {1}{2}\right )^{-2 n}+8 \operatorname {HeunCPrime}\left (0, -2 n +1, 0, 0, n^{2}-n +\frac {1}{2}, \frac {2}{x +1}\right ) c_{1} \right )\right )\right ) \left (-\frac {x}{2}-\frac {1}{2}\right )^{2 n}}{\left (x +1\right )^{2} n \left (8 c_{1} \operatorname {hypergeom}\left (\left [1-n , 1-n \right ], \left [-2 n +2\right ], -\frac {2}{x -1}\right ) \left (-\frac {x}{2}-\frac {1}{2}\right )^{2 n}+\left (\frac {x +1}{x -1}\right )^{2 n} \operatorname {hypergeom}\left (\left [n , n\right ], \left [2 n \right ], -\frac {2}{x -1}\right ) \left (x -1\right )\right )}
\]
✓ Solution by Mathematica
Time used: 0.342 (sec). Leaf size: 47
DSolve[(1-x^2)*D[y[x],x]==n*(1-2*x*y[x]+y[x]^2),y[x],x,IncludeSingularSolutions -> True]
\begin{align*}
y(x)\to \frac {\operatorname {LegendreQ}(n,x)+c_1 \operatorname {LegendreP}(n,x)}{\operatorname {LegendreQ}(n-1,x)+c_1 \operatorname {LegendreP}(n-1,x)} \\
y(x)\to \frac {\operatorname {LegendreP}(n,x)}{\operatorname {LegendreP}(n-1,x)} \\
\end{align*}