Internal problem ID [9754]
Internal file name [OUTPUT/8696_Monday_June_06_2022_05_14_09_AM_70763383/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1427.
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.
\[ \boxed {y^{\prime \prime }+\frac {\left (-\left (a^{2} b^{2}-\left (a +1\right )^{2}\right ) \sin \left (x \right )^{2}-a \left (a +1\right ) b \sin \left (2 x \right )-a \left (a -1\right )\right ) y}{\sin \left (x \right )^{2}}=0} \]
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 symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> 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 -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> 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 a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients 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 = arccot(t)] Linear ODE actually solved: (-a^2*b^2-2*a^2*b*t-a^2*t^2-2*a*b*t+a*t^2+3*a+1)*u(t)+(2*t^3+2*t)*diff(u(t),t)+(t^4+2*t^2+1)*diff(diff(u(t),t),t) = 0 <- change of variables successful`
✓ Solution by Maple
Time used: 0.594 (sec). Leaf size: 98
dsolve(diff(diff(y(x),x),x) = -(-(a^2*b^2-(a+1)^2)*sin(x)^2-a*(a+1)*b*sin(2*x)-a*(a-1))/sin(x)^2*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \left (c_{2} \left (\cot \left (x \right )+i\right )^{\frac {1}{2}+\frac {1}{2} a +\frac {1}{2} i a b} \operatorname {hypergeom}\left (\left [i a b -a +1, a \left (i b +1\right )\right ], \left [i a b +a +2\right ], \frac {1}{2}-\frac {i \cot \left (x \right )}{2}\right )+c_{1} \left (\cot \left (x \right )+i\right )^{-\frac {1}{2}-\frac {1}{2} i a b -\frac {1}{2} a} \left (b +\cot \left (x \right )\right )\right ) \left (\cot \left (x \right )-i\right )^{-\frac {1}{2}+\frac {1}{2} i a b -\frac {1}{2} a} \]
✓ Solution by Mathematica
Time used: 1.502 (sec). Leaf size: 161
DSolve[y''[x] == -(Csc[x]^2*((1 - a)*a - (-(1 + a)^2 + a^2*b^2)*Sin[x]^2 - a*(1 + a)*b*Sin[2*x])*y[x]),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 e^{-a b x} \sin ^{-a}(x) \left (\csc (x)+\frac {2^{2 a+1} (2 a+1) e^{2 i x} \left (1-e^{2 i x}\right )^{2 a} \left (-i e^{-i x} \left (-1+e^{2 i x}\right )\right )^{-2 a} \sin ^{2 a}(x) \operatorname {Hypergeometric2F1}\left (2 a+2,i b a+a+1,i b a+a+2,e^{2 i x}\right ) (b \sin (x)+\cos (x))}{a (b-i)-i}\right )+c_1 e^{a b x} \sin ^a(x) (b \sin (x)+\cos (x)) \]