56.2.11 problem 11
Internal
problem
ID
[8815]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
2.0
Problem
number
:
11
Date
solved
:
Tuesday, January 28, 2025 at 03:16:53 PM
CAS
classification
:
[[_2nd_order, _with_linear_symmetries]]
\begin{align*} y^{\prime \prime }-a x y^{\prime }-b x y-c x&=0 \end{align*}
✓ Solution by Maple
Time used: 0.026 (sec). Leaf size: 86
dsolve(diff(y(x),x$2)-a*x*diff(y(x),x)-b*x*y(x)-c*x=0,y(x), singsol=all)
\[
y = \frac {{\mathrm e}^{-\frac {b x}{a}} \operatorname {KummerM}\left (-\frac {b^{2}}{2 a^{3}}, \frac {1}{2}, \frac {\left (a^{2} x +2 b \right )^{2}}{2 a^{3}}\right ) c_{2} b +{\mathrm e}^{-\frac {b x}{a}} \operatorname {KummerU}\left (-\frac {b^{2}}{2 a^{3}}, \frac {1}{2}, \frac {\left (a^{2} x +2 b \right )^{2}}{2 a^{3}}\right ) c_{1} b -c}{b}
\]
✓ Solution by Mathematica
Time used: 4.917 (sec). Leaf size: 565
DSolve[D[y[x],{x,2}]-a*x*D[y[x],x]-b*x*y[x]-c*x==0,y[x],x,IncludeSingularSolutions -> True]
\[
y(x)\to e^{-\frac {b x}{a}} \left (\operatorname {HermiteH}\left (\frac {b^2}{a^3},\frac {x a^2+2 b}{\sqrt {2} a^{3/2}}\right ) \int _1^x\frac {a^4 c e^{\frac {b K[1]}{a}} \operatorname {Hypergeometric1F1}\left (-\frac {b^2}{2 a^3},\frac {1}{2},\frac {\left (K[1] a^2+2 b\right )^2}{2 a^3}\right ) K[1]}{b^2 \left (\sqrt {2} \operatorname {HermiteH}\left (\frac {b^2}{a^3}-1,\frac {K[1] a^2+2 b}{\sqrt {2} a^{3/2}}\right ) \operatorname {Hypergeometric1F1}\left (-\frac {b^2}{2 a^3},\frac {1}{2},\frac {\left (K[1] a^2+2 b\right )^2}{2 a^3}\right ) a^{3/2}+\operatorname {HermiteH}\left (\frac {b^2}{a^3},\frac {K[1] a^2+2 b}{\sqrt {2} a^{3/2}}\right ) \operatorname {Hypergeometric1F1}\left (1-\frac {b^2}{2 a^3},\frac {3}{2},\frac {\left (K[1] a^2+2 b\right )^2}{2 a^3}\right ) \left (K[1] a^2+2 b\right )\right )}dK[1]+\operatorname {Hypergeometric1F1}\left (-\frac {b^2}{2 a^3},\frac {1}{2},\frac {\left (x a^2+2 b\right )^2}{2 a^3}\right ) \int _1^x-\frac {a^4 c e^{\frac {b K[2]}{a}} \operatorname {HermiteH}\left (\frac {b^2}{a^3},\frac {K[2] a^2+2 b}{\sqrt {2} a^{3/2}}\right ) K[2]}{b^2 \left (\sqrt {2} \operatorname {HermiteH}\left (\frac {b^2}{a^3}-1,\frac {K[2] a^2+2 b}{\sqrt {2} a^{3/2}}\right ) \operatorname {Hypergeometric1F1}\left (-\frac {b^2}{2 a^3},\frac {1}{2},\frac {\left (K[2] a^2+2 b\right )^2}{2 a^3}\right ) a^{3/2}+\operatorname {HermiteH}\left (\frac {b^2}{a^3},\frac {K[2] a^2+2 b}{\sqrt {2} a^{3/2}}\right ) \operatorname {Hypergeometric1F1}\left (1-\frac {b^2}{2 a^3},\frac {3}{2},\frac {\left (K[2] a^2+2 b\right )^2}{2 a^3}\right ) \left (K[2] a^2+2 b\right )\right )}dK[2]+c_2 \operatorname {Hypergeometric1F1}\left (-\frac {b^2}{2 a^3},\frac {1}{2},\frac {\left (x a^2+2 b\right )^2}{2 a^3}\right )+c_1 \operatorname {HermiteH}\left (\frac {b^2}{a^3},\frac {x a^2+2 b}{\sqrt {2} a^{3/2}}\right )\right )
\]