29.30.8 problem 867
Internal
problem
ID
[5448]
Book
:
Ordinary
differential
equations
and
their
solutions.
By
George
Moseley
Murphy.
1960
Section
:
Various
30
Problem
number
:
867
Date
solved
:
Monday, January 27, 2025 at 11:24:12 AM
CAS
classification
:
[[_homogeneous, `class A`], _rational, _dAlembert]
\begin{align*} x {y^{\prime }}^{2}-2 y^{\prime } y+a x&=0 \end{align*}
✓ Solution by Maple
Time used: 0.088 (sec). Leaf size: 33
dsolve(x*diff(y(x),x)^2-2*y(x)*diff(y(x),x)+a*x = 0,y(x), singsol=all)
\begin{align*}
y \left (x \right ) &= x \sqrt {a} \\
y \left (x \right ) &= -x \sqrt {a} \\
y \left (x \right ) &= \frac {\left (\frac {x^{2}}{c_{1}^{2}}+a \right ) c_{1}}{2} \\
\end{align*}
✓ Solution by Mathematica
Time used: 19.313 (sec). Leaf size: 519
DSolve[x (D[y[x],x])^2-2 y[x] D[y[x],x]+a x==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*}
y(x)\to -\frac {\sqrt {a} x \tan (c_1-i \log (x))}{\sqrt {\sec ^2(c_1-i \log (x))}} \\
y(x)\to \frac {\sqrt {a} x \tan (c_1-i \log (x))}{\sqrt {\sec ^2(c_1-i \log (x))}} \\
y(x)\to -\frac {\sqrt {a} x \tan (i \log (x)+c_1)}{\sqrt {\sec ^2(i \log (x)+c_1)}} \\
y(x)\to \frac {\sqrt {a} x \tan (i \log (x)+c_1)}{\sqrt {\sec ^2(i \log (x)+c_1)}} \\
y(x)\to -\sqrt {a} x \\
y(x)\to \sqrt {a} x \\
y(x)\to \frac {i \sqrt {a} \left (e^{2 i \text {Interval}[\{0,2 \pi \}]} \sqrt {\frac {x^2 e^{2 i \text {Interval}[\{0,\pi \}]}}{\left (x^2+e^{2 i \text {Interval}[\{0,\pi \}]}\right )^2}}-x^4 e^{2 i \text {Interval}[\{0,\pi \}]} \sqrt {\frac {x^2 e^{2 i \text {Interval}[\{0,\pi \}]}}{\left (x^2+e^{2 i \text {Interval}[\{0,\pi \}]}\right )^2}}\right )}{2 x} \\
y(x)\to \frac {i \sqrt {a} \left (x^4 e^{2 i \text {Interval}[\{0,\pi \}]} \sqrt {\frac {x^2 e^{2 i \text {Interval}[\{0,\pi \}]}}{\left (x^2+e^{2 i \text {Interval}[\{0,\pi \}]}\right )^2}}-e^{2 i \text {Interval}[\{0,2 \pi \}]} \sqrt {\frac {x^2 e^{2 i \text {Interval}[\{0,\pi \}]}}{\left (x^2+e^{2 i \text {Interval}[\{0,\pi \}]}\right )^2}}\right )}{2 x} \\
y(x)\to \frac {i \sqrt {a} \left (x^4 e^{2 i \text {Interval}[\{0,\pi \}]} \sqrt {\frac {x^2 e^{2 i \text {Interval}[\{0,\pi \}]}}{\left (x^2+e^{2 i \text {Interval}[\{0,\pi \}]}\right )^2}}-e^{2 i \text {Interval}[\{0,2 \pi \}]} \sqrt {\frac {x^2 e^{2 i \text {Interval}[\{0,\pi \}]}}{\left (x^2+e^{2 i \text {Interval}[\{0,\pi \}]}\right )^2}}\right )}{2 x} \\
\end{align*}