Internal problem ID [6554]
Book: Own collection of miscellaneous problems
Section: section 5.0
Problem number: 15.
ODE order: 1.
ODE degree: 2.
CAS Maple gives this as type [_quadrature]
\[ \boxed {h^{2}+\frac {2 a h}{\sqrt {1+{h^{\prime }}^{2}}}-b^{2}=0} \]
✓ Solution by Maple
Time used: 0.719 (sec). Leaf size: 108
dsolve(h(u)^2 + 2*a*h(u)/sqrt(1 + diff(h(u), u)^2) = b^2,h(u), singsol=all)
\begin{align*} u -\left (\int _{}^{h \left (u \right )}\frac {\left (\textit {\_a} +b \right ) \left (\textit {\_a} -b \right )}{\sqrt {-\left (\textit {\_a}^{2}+2 \textit {\_a} a -b^{2}\right ) \left (\textit {\_a}^{2}-2 \textit {\_a} a -b^{2}\right )}}d \textit {\_a} \right )-c_{1} = 0 \\ u -\left (\int _{}^{h \left (u \right )}-\frac {\left (\textit {\_a} +b \right ) \left (\textit {\_a} -b \right )}{\sqrt {-\left (\textit {\_a}^{2}+2 \textit {\_a} a -b^{2}\right ) \left (\textit {\_a}^{2}-2 \textit {\_a} a -b^{2}\right )}}d \textit {\_a} \right )-c_{1} = 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 24.533 (sec). Leaf size: 913
DSolve[h[u]^2 + 2*a*h[u]/Sqrt[1 + (h'[u])^2] == b^2,h[u],u,IncludeSingularSolutions -> True]
\begin{align*} h(u)\to \text {InverseFunction}\left [-\frac {i \sqrt {\left (b^2-\text {$\#$1}^2\right )^2} \sqrt {1-\frac {\text {$\#$1}^2}{-2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2}} \sqrt {1-\frac {\text {$\#$1}^2}{2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2}} \left (\left (2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2\right ) E\left (i \text {arcsinh}\left (\sqrt {\frac {1}{-2 a^2-b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}} \text {$\#$1}\right )|\frac {2 a^2+b^2-2 \sqrt {a^2 \left (a^2+b^2\right )}}{2 a^2+b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}\right )-2 \left (\sqrt {a^2 \left (a^2+b^2\right )}+a^2\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {1}{-2 a^2-b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}} \text {$\#$1}\right ),\frac {2 a^2+b^2-2 \sqrt {a^2 \left (a^2+b^2\right )}}{2 a^2+b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}\right )\right )}{\left (b^2-\text {$\#$1}^2\right ) \sqrt {\frac {1}{2 \sqrt {a^2 \left (a^2+b^2\right )}-2 a^2-b^2}} \sqrt {-\text {$\#$1}^4+4 \text {$\#$1}^2 a^2+2 \text {$\#$1}^2 b^2-b^4}}\&\right ][-u+c_1] \\ h(u)\to \text {InverseFunction}\left [-\frac {i \sqrt {\left (b^2-\text {$\#$1}^2\right )^2} \sqrt {1-\frac {\text {$\#$1}^2}{-2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2}} \sqrt {1-\frac {\text {$\#$1}^2}{2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2}} \left (\left (2 \sqrt {a^2 \left (a^2+b^2\right )}+2 a^2+b^2\right ) E\left (i \text {arcsinh}\left (\sqrt {\frac {1}{-2 a^2-b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}} \text {$\#$1}\right )|\frac {2 a^2+b^2-2 \sqrt {a^2 \left (a^2+b^2\right )}}{2 a^2+b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}\right )-2 \left (\sqrt {a^2 \left (a^2+b^2\right )}+a^2\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {1}{-2 a^2-b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}} \text {$\#$1}\right ),\frac {2 a^2+b^2-2 \sqrt {a^2 \left (a^2+b^2\right )}}{2 a^2+b^2+2 \sqrt {a^2 \left (a^2+b^2\right )}}\right )\right )}{\left (b^2-\text {$\#$1}^2\right ) \sqrt {\frac {1}{2 \sqrt {a^2 \left (a^2+b^2\right )}-2 a^2-b^2}} \sqrt {-\text {$\#$1}^4+4 \text {$\#$1}^2 a^2+2 \text {$\#$1}^2 b^2-b^4}}\&\right ][u+c_1] \\ h(u)\to -\sqrt {a^2+b^2}-a \\ h(u)\to \sqrt {a^2+b^2}-a \\ \end{align*}