Internal problem ID [10131]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1809 (book 6.218).
ODE order: 2.
ODE degree: 1.
CAS Maple gives this as type
[[_2nd_order, _missing_x], _Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {\left (y^{2}-1\right ) \left (a^{2} y^{2}-1\right ) y^{\prime \prime }+b \sqrt {\left (1-y^{2}\right ) \left (1-a^{2} y^{2}\right )}\, {y^{\prime }}^{2}+\left (1+a^{2}-2 a^{2} y^{2}\right ) y {y^{\prime }}^{2}=0} \]
✗ Solution by Maple
dsolve((y(x)^2-1)*(a^2*y(x)^2-1)*diff(diff(y(x),x),x)+b*((1-y(x)^2)*(1-a^2*y(x)^2))^(1/2)*diff(y(x),x)^2+(1+a^2-2*a^2*y(x)^2)*y(x)*diff(y(x),x)^2=0,y(x), singsol=all)
\[ \text {No solution found} \]
✓ Solution by Mathematica
Time used: 22.452 (sec). Leaf size: 372
DSolve[y[x]*(1 + a^2 - 2*a^2*y[x]^2)*y'[x]^2 + b*Sqrt[(1 - y[x]^2)*(1 - a^2*y[x]^2)]*y'[x]^2 + (-1 + y[x]^2)*(-1 + a^2*y[x]^2)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {\exp \left (\frac {b \sqrt {1-K[1]^2} \sqrt {1-a^2 K[1]^2} \operatorname {EllipticF}\left (\arcsin (K[1]),a^2\right )}{\sqrt {\left (K[1]^2-1\right ) \left (a^2 K[1]^2-1\right )}}+\frac {1}{2} (-\log (1-K[1])-\log (K[1]+1)-\log (1-a K[1])-\log (a K[1]+1))\right )}{c_1}dK[1]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}-\frac {\exp \left (\frac {b \sqrt {1-K[1]^2} \sqrt {1-a^2 K[1]^2} \operatorname {EllipticF}\left (\arcsin (K[1]),a^2\right )}{\sqrt {\left (K[1]^2-1\right ) \left (a^2 K[1]^2-1\right )}}+\frac {1}{2} (-\log (1-K[1])-\log (K[1]+1)-\log (1-a K[1])-\log (a K[1]+1))\right )}{c_1}dK[1]\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {\exp \left (\frac {b \sqrt {1-K[1]^2} \sqrt {1-a^2 K[1]^2} \operatorname {EllipticF}\left (\arcsin (K[1]),a^2\right )}{\sqrt {\left (K[1]^2-1\right ) \left (a^2 K[1]^2-1\right )}}+\frac {1}{2} (-\log (1-K[1])-\log (K[1]+1)-\log (1-a K[1])-\log (a K[1]+1))\right )}{c_1}dK[1]\&\right ][x+c_2] \\ \end{align*}