Internal problem ID [10137]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1815 (book 6.224).
ODE order: 2.
ODE degree: 0.
CAS Maple gives this as type [[_2nd_order, _missing_x]]
\[ \boxed {h \left (y\right ) y^{\prime \prime }+a D\left (h \right )\left (y\right ) {y^{\prime }}^{2}+j \left (y\right )=0} \]
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 79
dsolve(h(y(x))*diff(diff(y(x),x),x)+a*D(h)(y(x))*diff(y(x),x)^2+j(y(x))=0,y(x), singsol=all)
\begin{align*} \int _{}^{y \left (x \right )}\frac {h \left (\textit {\_b} \right )^{a}}{\sqrt {-2 \left (\int j \left (\textit {\_b} \right ) h \left (\textit {\_b} \right )^{-1+2 a}d \textit {\_b} \right )+c_{1}}}d \textit {\_b} -x -c_{2} &= 0 \\ -\left (\int _{}^{y \left (x \right )}\frac {h \left (\textit {\_b} \right )^{a}}{\sqrt {-2 \left (\int j \left (\textit {\_b} \right ) h \left (\textit {\_b} \right )^{-1+2 a}d \textit {\_b} \right )+c_{1}}}d \textit {\_b} \right )-x -c_{2} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.935 (sec). Leaf size: 362
DSolve[j[y[x]] + a*h[y[x]]*y'[x]^2 + h[y[x]]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}-\frac {e^{a K[2]}}{\sqrt {c_1+2 \int _1^{K[2]}-\frac {e^{2 a K[1]} j(K[1])}{h(K[1])}dK[1]}}dK[2]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {e^{a K[3]}}{\sqrt {c_1+2 \int _1^{K[3]}-\frac {e^{2 a K[1]} j(K[1])}{h(K[1])}dK[1]}}dK[3]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}-\frac {e^{a K[2]}}{\sqrt {2 \int _1^{K[2]}-\frac {e^{2 a K[1]} j(K[1])}{h(K[1])}dK[1]-c_1}}dK[2]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}-\frac {e^{a K[2]}}{\sqrt {c_1+2 \int _1^{K[2]}-\frac {e^{2 a K[1]} j(K[1])}{h(K[1])}dK[1]}}dK[2]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {e^{a K[3]}}{\sqrt {2 \int _1^{K[3]}-\frac {e^{2 a K[1]} j(K[1])}{h(K[1])}dK[1]-c_1}}dK[3]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {e^{a K[3]}}{\sqrt {c_1+2 \int _1^{K[3]}-\frac {e^{2 a K[1]} j(K[1])}{h(K[1])}dK[1]}}dK[3]\&\right ][x+c_2] \\ \end{align*}