4.80 problem 1530

Internal problem ID [9854]
Internal file name [OUTPUT/8799_Monday_June_06_2022_05_30_36_AM_99705687/index.tex]

Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 3, linear third order
Problem number: 1530.
ODE order: 3.
ODE degree: 1.

The type(s) of ODE detected by this program : "unknown"

Maple gives the following as the ode type

[[_3rd_order, _with_linear_symmetries]]

Unable to solve or complete the solution.

\[ \boxed {y^{\prime \prime \prime } \sin \left (x \right )^{2}+3 y^{\prime \prime } \sin \left (x \right ) \cos \left (x \right )+\left (\cos \left (2 x \right )+4 \nu \left (\nu +1\right ) \sin \left (x \right )^{2}\right ) y^{\prime }+2 \nu \left (\nu +1\right ) y \sin \left (2 x \right )=0} \] Unable to solve this ODE.

4.80.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime \prime }\right ) \sin \left (x \right )^{2}+3 \left (\frac {d}{d x}y^{\prime }\right ) \sin \left (x \right ) \cos \left (x \right )+\left (\cos \left (2 x \right )+4 \nu \left (\nu +1\right ) \sin \left (x \right )^{2}\right ) y^{\prime }+2 \nu \left (\nu +1\right ) y \sin \left (2 x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \frac {d}{d x}y^{\prime \prime } \end {array} \]

Maple trace

`Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
trying high order exact linear fully integrable 
trying to convert to a linear ODE with constant coefficients 
trying differential order: 3; missing the dependent variable 
Equation is the 2nd symmetric power of diff(diff(y(x),x),x)+cos(x)/sin(x)*diff(y(x),x)-1/4*(-4*sin(x)^2*nu^2-4*sin(x)^2*nu+cos(x)^2- 
-> Attempting now to solve this lower order ODE 
   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: 
         (-nu^2-nu)*u(t)+(6*t+2)*diff(u(t),t)+(4*t^2-4)*diff(diff(u(t),t),t) = 0 
   <- change of variables successful`
 

✓ Solution by Maple

Time used: 0.36 (sec). Leaf size: 105

dsolve(diff(diff(diff(y(x),x),x),x)*sin(x)^2+3*diff(diff(y(x),x),x)*sin(x)*cos(x)+(cos(2*x)+4*nu*(nu+1)*sin(x)^2)*diff(y(x),x)+2*nu*(nu+1)*y(x)*sin(2*x)=0,y(x), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.086 (sec). Leaf size: 35

DSolve[2*nu*(1 + nu)*Sin[2*x]*y[x] + (Cos[2*x] + 4*nu*(1 + nu)*Sin[x]^2)*y'[x] + 3*Cos[x]*Sin[x]*y''[x] + Sin[x]^2*Derivative[3][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to c_3 \operatorname {LegendreP}(\nu ,\cos (x)) \operatorname {LegendreQ}(\nu ,\cos (x))+c_1 \operatorname {LegendreP}(\nu ,\cos (x))^2+c_2 \operatorname {LegendreQ}(\nu ,\cos (x))^2 \]