2.52 problem 51

Internal problem ID [7836]
Book : Own collection of miscellaneous problems
Section : section 2.0
Problem number : 51
Date solved : Tuesday, October 22, 2024 at 02:46:50 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*}

2.52.1 Maple step by step solution

2.52.2 Maple trace

Methods for second order ODEs:
 

2.52.3 Maple dsolve solution

Solving time : 0.029 (sec)
Leaf size : 28

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

\[ y = x \left (\operatorname {KummerU}\left (\frac {1}{2}, \frac {5}{4}, \frac {x^{4}}{4}\right ) c_1 +\operatorname {KummerM}\left (\frac {1}{2}, \frac {5}{4}, \frac {x^{4}}{4}\right ) c_2 -\frac {1}{2}\right ) \]

2.52.4 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]
 

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