60.7.140 problem 1731 (book 6.140)
Internal
problem
ID
[11729]
Book
:
Differential
Gleichungen,
E.
Kamke,
3rd
ed.
Chelsea
Pub.
NY,
1948
Section
:
Chapter
6,
non-linear
second
order
Problem
number
:
1731
(book
6.140)
Date
solved
:
Monday, January 27, 2025 at 11:31:57 PM
CAS
classification
:
[[_2nd_order, _missing_x]]
\begin{align*} 2 y^{\prime \prime } y-{y^{\prime }}^{2}-8 y^{3}&=0 \end{align*}
✓ Solution by Maple
Time used: 0.082 (sec). Leaf size: 57
dsolve(2*diff(diff(y(x),x),x)*y(x)-diff(y(x),x)^2-8*y(x)^3=0,y(x), singsol=all)
\begin{align*}
y &= 0 \\
\int _{}^{y}\frac {1}{\sqrt {\textit {\_a} \left (4 \textit {\_a}^{2}+c_{1} \right )}}d \textit {\_a} -x -c_{2} &= 0 \\
-\int _{}^{y}\frac {1}{\sqrt {\textit {\_a} \left (4 \textit {\_a}^{2}+c_{1} \right )}}d \textit {\_a} -x -c_{2} &= 0 \\
\end{align*}
✓ Solution by Mathematica
Time used: 0.979 (sec). Leaf size: 415
DSolve[-8*y[x]^3 - D[y[x],x]^2 + 2*y[x]*D[y[x],{x,2}] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*}
y(x)\to \text {InverseFunction}\left [-\frac {2 \sqrt {\text {$\#$1}} \sqrt {1+\frac {4 \text {$\#$1}^2}{c_1}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {4 \text {$\#$1}^2}{c_1}\right )}{\sqrt {4 \text {$\#$1}^2+c_1}}\&\right ][x+c_2] \\
y(x)\to \text {InverseFunction}\left [\frac {2 \sqrt {\text {$\#$1}} \sqrt {1+\frac {4 \text {$\#$1}^2}{c_1}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {4 \text {$\#$1}^2}{c_1}\right )}{\sqrt {4 \text {$\#$1}^2+c_1}}\&\right ][x+c_2] \\
y(x)\to \text {InverseFunction}\left [-\frac {2 \sqrt {\text {$\#$1}} \sqrt {1-\frac {4 \text {$\#$1}^2}{c_1}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {-4 \text {$\#$1}^2}{-c_1}\right )}{\sqrt {4 \text {$\#$1}^2-c_1}}\&\right ][x+c_2] \\
y(x)\to \text {InverseFunction}\left [\frac {2 \sqrt {\text {$\#$1}} \sqrt {1-\frac {4 \text {$\#$1}^2}{c_1}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {-4 \text {$\#$1}^2}{-c_1}\right )}{\sqrt {4 \text {$\#$1}^2-c_1}}\&\right ][x+c_2] \\
y(x)\to \text {InverseFunction}\left [-\frac {2 \sqrt {\text {$\#$1}} \sqrt {1+\frac {4 \text {$\#$1}^2}{c_1}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {4 \text {$\#$1}^2}{c_1}\right )}{\sqrt {4 \text {$\#$1}^2+c_1}}\&\right ][x+c_2] \\
y(x)\to \text {InverseFunction}\left [\frac {2 \sqrt {\text {$\#$1}} \sqrt {1+\frac {4 \text {$\#$1}^2}{c_1}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {4 \text {$\#$1}^2}{c_1}\right )}{\sqrt {4 \text {$\#$1}^2+c_1}}\&\right ][x+c_2] \\
\end{align*}