problem 1425

Internal problem ID [9752]
Internal ﬁle name [OUTPUT/8694_Monday_June_06_2022_05_13_44_AM_48608672/index.tex]

Book: Diﬀerential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1425.
ODE order: 2.
ODE degree: 1.

\[ \boxed {y^{\prime \prime }+\frac {\left (-a^{2} \cos \left (x \right )^{2}-\left (3-2 a \right ) \cos \left (x \right )-3+3 a \right ) y}{\sin \left (x \right )^{2}}=0} \]

Maple trace Kovacic algorithm successful

`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 
      A Liouvillian solution exists 
      Reducible group (found an exponential solution) 
      Group is reducible, not completely reducible 
      Solution has integrals. Trying a special function solution free of integrals... 
      -> Trying a solution in terms of special functions: 
         -> Bessel 
         -> elliptic 
         -> Legendre 
         -> Whittaker 
            -> 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 
         -> Trying to convert hypergeometric functions to elementary form... 
         <- elementary form for at least one hypergeometric solution is achieved - returning with no uncomputed integrals 
      <- Kovacics algorithm successful 
   Change of variables used: 
      [x = arccos(t)] 
   Linear ODE actually solved: 
      (-2*a^2*t^2+4*a*t+6*a-6*t-6)*u(t)+(2*t^3-2*t)*diff(u(t),t)+(2*t^4-4*t^2+2)*diff(diff(u(t),t),t) = 0 
<- change of variables successful`

\[ y \left (x \right ) = \frac {c_{1} \left (-2+\left (2 a -1\right ) \cos \left (x \right )\right ) \sqrt {\cos \left (\frac {x}{2}\right )}\, \sin \left (x \right )^{a -\frac {1}{2}}}{\sin \left (\frac {x}{2}\right )^{\frac {3}{2}}}+\frac {c_{2} \left (\frac {\cos \left (x \right )}{2}-\frac {1}{2}\right )^{\frac {a}{2}-\frac {3}{4}} \left (\frac {\cos \left (x \right )}{2}+\frac {1}{2}\right )^{\frac {3}{4}-\frac {a}{2}} \operatorname {hypergeom}\left (\left [a -\frac {1}{2}, -a -\frac {1}{2}\right ], \left [\frac {3}{2}-a \right ], \frac {\cos \left (x \right )}{2}+\frac {1}{2}\right )}{\sqrt {\sin \left (x \right )}} \]

\[ y(x)\to \frac {c_1 \sin ^2(x)^{a/2} (-2 a \cos (x)+\cos (x)+2)}{1-\cos (x)}-\frac {c_2 \sin ^2(x)^{-a} (1-\cos (x))^{\frac {a-1}{2}} (\cos (x)+1)^{\frac {a+1}{2}} \left (\frac {(2 a-1) (\cos (x)-1)}{(2 a-1) \cos (x)-2}\right )^{a-\frac {1}{2}} \left (\frac {(2 a-1) (\cos (x)+1)}{(2 a-1) \cos (x)-2}\right )^{a-\frac {1}{2}} \operatorname {AppellF1}\left (2 a,a-\frac {3}{2},a+\frac {1}{2},2 a+1,\frac {3-2 a}{-2 a \cos (x)+\cos (x)+2},\frac {2 a+1}{-2 a \cos (x)+\cos (x)+2}\right )}{4 a^2-2 a} \]

3.419 problem 1425