Internal problem ID [9763]
Internal file name [OUTPUT/8705_Monday_June_06_2022_05_16_07_AM_37193491/index.tex
]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1436.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime \prime }+\frac {\left (4 v \left (v +1\right ) \sin \left (x \right )^{2}-\cos \left (x \right )^{2}+2-4 n^{2}\right ) y}{4 \sin \left (x \right )^{2}}=0} \]
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 -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 2F1 ODE <- hypergeometric successful <- special function solution successful Change of variables used: [x = 1/2*arccos(t)] Linear ODE actually solved: (-4*t*v^2-8*n^2-4*t*v+4*v^2-t+4*v+3)*u(t)+(16*t^2-16*t)*diff(u(t),t)+(16*t^3-16*t^2-16*t+16)*diff(diff(u(t),t),t) = 0 <- change of variables successful`
✓ Solution by Maple
Time used: 0.578 (sec). Leaf size: 91
dsolve(diff(diff(y(x),x),x) = -1/4*(4*v*(v+1)*sin(x)^2-cos(x)^2+2-4*n^2)/sin(x)^2*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\sqrt {\cos \left (x \right )}\, \left (-\frac {1}{2}+\frac {\cos \left (2 x \right )}{2}\right )^{\frac {n}{2}+\frac {1}{2}} \left (c_{1} \operatorname {hypergeom}\left (\left [-\frac {v}{2}+\frac {n}{2}, \frac {1}{2}+\frac {v}{2}+\frac {n}{2}\right ], \left [\frac {1}{2}\right ], \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )+c_{2} \cos \left (x \right ) \operatorname {hypergeom}\left (\left [1+\frac {v}{2}+\frac {n}{2}, \frac {1}{2}-\frac {v}{2}+\frac {n}{2}\right ], \left [\frac {3}{2}\right ], \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )\right )}{\sqrt {\sin \left (2 x \right )}} \]
✓ Solution by Mathematica
Time used: 0.66 (sec). Leaf size: 33
DSolve[y''[x] == -1/4*(Csc[x]^2*(2 - 4*n^2 - Cos[x]^2 + 4*v*(1 + v)*Sin[x]^2)*y[x]),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \sqrt [4]{-\sin ^2(x)} (c_1 P_v^n(\cos (x))+c_2 Q_v^n(\cos (x))) \]