2.2.52 problem 51

Internal problem ID [8282]
Book : Own collection of miscellaneous problems
Section : section 2.0
Problem number : 51
Date solved : Sunday, November 10, 2024 at 09:09:42 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

\begin{align*} y^{\prime \prime }-x^{3} y^{\prime }-x^{2} y-x^{3}&=0 \end{align*}

Maple step by step solution

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
-> Try solving first the homogeneous part of the ODE 
   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 
         <- hyper3 successful: received ODE is equivalent to the 1F1 ODE 
      <- Kummer successful 
   <- special function solution successful 
<- solving first the homogeneous part of the ODE successful`
 

Maple dsolve solution

Solving time : 0.216 (sec)
Leaf size : 28

dsolve(diff(diff(y(x),x),x)-diff(y(x),x)*x^3-x^2*y(x)-x^3 = 0, 
       y(x),singsol=all)
 

\[ y = \left (\operatorname {KummerU}\left (\frac {1}{2}, \frac {5}{4}, \frac {x^{4}}{4}\right ) c_{1} +\operatorname {KummerM}\left (\frac {1}{2}, \frac {5}{4}, \frac {x^{4}}{4}\right ) c_{2} -\frac {1}{2}\right ) x \]

Mathematica DSolve solution

Solving time : 1.16 (sec)
Leaf size : 337

DSolve[{D[y[x],{x,2}]-x^3*D[y[x],x]-x^2*y[x]-x^3==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to \operatorname {Hypergeometric1F1}\left (\frac {1}{4},\frac {3}{4},\frac {x^4}{4}\right ) \int _1^x\frac {15 \operatorname {Hypergeometric1F1}\left (\frac {1}{2},\frac {5}{4},\frac {K[1]^4}{4}\right ) K[1]^4}{5 \operatorname {Hypergeometric1F1}\left (\frac {1}{2},\frac {5}{4},\frac {K[1]^4}{4}\right ) \operatorname {Hypergeometric1F1}\left (\frac {5}{4},\frac {7}{4},\frac {K[1]^4}{4}\right ) K[1]^4-3 \operatorname {Hypergeometric1F1}\left (\frac {1}{4},\frac {3}{4},\frac {K[1]^4}{4}\right ) \left (2 \operatorname {Hypergeometric1F1}\left (\frac {3}{2},\frac {9}{4},\frac {K[1]^4}{4}\right ) K[1]^4+5 \operatorname {Hypergeometric1F1}\left (\frac {1}{2},\frac {5}{4},\frac {K[1]^4}{4}\right )\right )}dK[1]+\frac {\sqrt [4]{-1} x \operatorname {Hypergeometric1F1}\left (\frac {1}{2},\frac {5}{4},\frac {x^4}{4}\right ) \int _1^x\frac {(15-15 i) \operatorname {Hypergeometric1F1}\left (\frac {1}{4},\frac {3}{4},\frac {K[2]^4}{4}\right ) K[2]^3}{3 \operatorname {Hypergeometric1F1}\left (\frac {1}{4},\frac {3}{4},\frac {K[2]^4}{4}\right ) \left (2 \operatorname {Hypergeometric1F1}\left (\frac {3}{2},\frac {9}{4},\frac {K[2]^4}{4}\right ) K[2]^4+5 \operatorname {Hypergeometric1F1}\left (\frac {1}{2},\frac {5}{4},\frac {K[2]^4}{4}\right )\right )-5 \operatorname {Hypergeometric1F1}\left (\frac {1}{2},\frac {5}{4},\frac {K[2]^4}{4}\right ) \operatorname {Hypergeometric1F1}\left (\frac {5}{4},\frac {7}{4},\frac {K[2]^4}{4}\right ) K[2]^4}dK[2]}{\sqrt {2}}+c_1 \operatorname {Hypergeometric1F1}\left (\frac {1}{4},\frac {3}{4},\frac {x^4}{4}\right )+\left (\frac {1}{2}+\frac {i}{2}\right ) c_2 x \operatorname {Hypergeometric1F1}\left (\frac {1}{2},\frac {5}{4},\frac {x^4}{4}\right ) \]