Internal problem ID [9312]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1733.
ODE order: 2.
ODE degree: 1.
CAS Maple gives this as type [[_2nd_order, _missing_x]]
Solve \begin {gather*} \boxed {2 y^{\prime \prime } y-\left (y^{\prime }\right )^{2}+\left (a y+b \right ) y^{2}=0} \end {gather*}
✓ Solution by Maple
Time used: 0.074 (sec). Leaf size: 75
dsolve(2*diff(diff(y(x),x),x)*y(x)-diff(y(x),x)^2+(a*y(x)+b)*y(x)^2=0,y(x), singsol=all)
\begin{align*} y \relax (x ) = 0 \\ \int _{}^{y \relax (x )}-\frac {2}{\sqrt {-2 \textit {\_a}^{3} a -4 b \,\textit {\_a}^{2}+4 \textit {\_a} c_{1}}}d \textit {\_a} -x -c_{2} = 0 \\ \int _{}^{y \relax (x )}\frac {2}{\sqrt {-2 \textit {\_a}^{3} a -4 b \,\textit {\_a}^{2}+4 \textit {\_a} c_{1}}}d \textit {\_a} -x -c_{2} = 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 1.268 (sec). Leaf size: 437
DSolve[y[x]^2*(b + a*y[x]) - y'[x]^2 + 2*y[x]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}\left [-\frac {i \sqrt {2} \text {$\#$1}^{3/2} \sqrt {2+\frac {4 c_1}{\text {$\#$1} \left (-b+\sqrt {b^2+2 a c_1}\right )}} \sqrt {1-\frac {2 c_1}{\text {$\#$1} \left (b+\sqrt {b^2+2 a c_1}\right )}} F\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c_1}{\sqrt {b^2+2 a c_1}-b}}}{\sqrt {\text {$\#$1}}}\right )|\frac {b-\sqrt {b^2+2 a c_1}}{b+\sqrt {b^2+2 a c_1}}\right )}{\sqrt {\frac {c_1}{-b+\sqrt {b^2+2 a c_1}}} \sqrt {-\text {$\#$1} \left (\text {$\#$1}^2 a+2 \text {$\#$1} b-2 c_1\right )}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {i \sqrt {2} \text {$\#$1}^{3/2} \sqrt {2+\frac {4 c_1}{\text {$\#$1} \left (-b+\sqrt {b^2+2 a c_1}\right )}} \sqrt {1-\frac {2 c_1}{\text {$\#$1} \left (b+\sqrt {b^2+2 a c_1}\right )}} F\left (i \sinh ^{-1}\left (\frac {\sqrt {2} \sqrt {\frac {c_1}{\sqrt {b^2+2 a c_1}-b}}}{\sqrt {\text {$\#$1}}}\right )|\frac {b-\sqrt {b^2+2 a c_1}}{b+\sqrt {b^2+2 a c_1}}\right )}{\sqrt {\frac {c_1}{-b+\sqrt {b^2+2 a c_1}}} \sqrt {-\text {$\#$1} \left (\text {$\#$1}^2 a+2 \text {$\#$1} b-2 c_1\right )}}\&\right ][x+c_2] \\ \end{align*}