56.2.52 problem 51
Internal
problem
ID
[8856]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
2.0
Problem
number
:
51
Date
solved
:
Tuesday, January 28, 2025 at 03:37:03 PM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
\begin{align*} y^{\prime \prime }-x^{3} y^{\prime }-x^{2} y-x^{3}&=0 \end{align*}
✓ Solution by Maple
Time used: 0.142 (sec). Leaf size: 28
dsolve(diff(y(x),x$2)-x^3*diff(y(x),x)-x^2*y(x)-x^3=0,y(x), singsol=all)
\[
y = \left (\operatorname {KummerU}\left (\frac {1}{2}, \frac {5}{4}, \frac {x^{4}}{4}\right ) c_{1} +\operatorname {KummerM}\left (\frac {1}{2}, \frac {5}{4}, \frac {x^{4}}{4}\right ) c_{2} -\frac {1}{2}\right ) x
\]
✓ Solution by Mathematica
Time used: 1.105 (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]
\[
y(x)\to \operatorname {Hypergeometric1F1}\left (\frac {1}{4},\frac {3}{4},\frac {x^4}{4}\right ) \int _1^x\frac {15 \operatorname {Hypergeometric1F1}\left (\frac {1}{2},\frac {5}{4},\frac {K[1]^4}{4}\right ) K[1]^4}{5 \operatorname {Hypergeometric1F1}\left (\frac {1}{2},\frac {5}{4},\frac {K[1]^4}{4}\right ) \operatorname {Hypergeometric1F1}\left (\frac {5}{4},\frac {7}{4},\frac {K[1]^4}{4}\right ) K[1]^4-3 \operatorname {Hypergeometric1F1}\left (\frac {1}{4},\frac {3}{4},\frac {K[1]^4}{4}\right ) \left (2 \operatorname {Hypergeometric1F1}\left (\frac {3}{2},\frac {9}{4},\frac {K[1]^4}{4}\right ) K[1]^4+5 \operatorname {Hypergeometric1F1}\left (\frac {1}{2},\frac {5}{4},\frac {K[1]^4}{4}\right )\right )}dK[1]+\frac {\sqrt [4]{-1} x \operatorname {Hypergeometric1F1}\left (\frac {1}{2},\frac {5}{4},\frac {x^4}{4}\right ) \int _1^x\frac {(15-15 i) \operatorname {Hypergeometric1F1}\left (\frac {1}{4},\frac {3}{4},\frac {K[2]^4}{4}\right ) K[2]^3}{3 \operatorname {Hypergeometric1F1}\left (\frac {1}{4},\frac {3}{4},\frac {K[2]^4}{4}\right ) \left (2 \operatorname {Hypergeometric1F1}\left (\frac {3}{2},\frac {9}{4},\frac {K[2]^4}{4}\right ) K[2]^4+5 \operatorname {Hypergeometric1F1}\left (\frac {1}{2},\frac {5}{4},\frac {K[2]^4}{4}\right )\right )-5 \operatorname {Hypergeometric1F1}\left (\frac {1}{2},\frac {5}{4},\frac {K[2]^4}{4}\right ) \operatorname {Hypergeometric1F1}\left (\frac {5}{4},\frac {7}{4},\frac {K[2]^4}{4}\right ) K[2]^4}dK[2]}{\sqrt {2}}+c_1 \operatorname {Hypergeometric1F1}\left (\frac {1}{4},\frac {3}{4},\frac {x^4}{4}\right )+\left (\frac {1}{2}+\frac {i}{2}\right ) c_2 x \operatorname {Hypergeometric1F1}\left (\frac {1}{2},\frac {5}{4},\frac {x^4}{4}\right )
\]