7.61 problem 1651

Internal problem ID [9230]

Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1651.
ODE order: 2.
ODE degree: 1.

CAS Maple gives this as type [[_2nd_order, _missing_x]]

Solve \begin {gather*} \boxed {y^{\prime \prime }-a \sqrt {\left (y^{\prime }\right )^{2}+1}-b=0} \end {gather*}

Solution by Maple

Time used: 0.075 (sec). Leaf size: 31

dsolve(diff(diff(y(x),x),x)-a*(diff(y(x),x)^2+1)^(1/2)-b=0,y(x), singsol=all)
 

\[ y \relax (x ) = \int \RootOf \left (x -\left (\int _{}^{\textit {\_Z}}\frac {1}{a \sqrt {\textit {\_f}^{2}+1}+b}d \textit {\_f} \right )+c_{1}\right )d x +c_{2} \]

Solution by Mathematica

Time used: 0.26 (sec). Leaf size: 769

DSolve[-b - a*Sqrt[1 + y'[x]^2] + y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to \frac {2 a \left (\sqrt {1+\text {InverseFunction}\left [\frac {\frac {2 b \text {ArcTan}\left (\frac {\left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right ) a+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\tanh ^{-1}\left (\frac {\text {$\#$1}}{\sqrt {\text {$\#$1}^2+1}}\right )}{a}\&\right ][x+c_1]{}^2}+a c_2\right )-b \left (\log \left (a^2 \text {InverseFunction}\left [\frac {\frac {2 b \text {ArcTan}\left (\frac {\left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right ) a+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\tanh ^{-1}\left (\frac {\text {$\#$1}}{\sqrt {\text {$\#$1}^2+1}}\right )}{a}\&\right ][x+c_1]{}^2+a^2-b^2\right )+\log \left (-\frac {2 a^2 \left (-i \sqrt {(a-b) (a+b)} \text {InverseFunction}\left [\frac {\frac {2 b \text {ArcTan}\left (\frac {\left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right ) a+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\tanh ^{-1}\left (\frac {\text {$\#$1}}{\sqrt {\text {$\#$1}^2+1}}\right )}{a}\&\right ][x+c_1]+b \sqrt {1+\text {InverseFunction}\left [\frac {\frac {2 b \text {ArcTan}\left (\frac {\left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right ) a+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\tanh ^{-1}\left (\frac {\text {$\#$1}}{\sqrt {\text {$\#$1}^2+1}}\right )}{a}\&\right ][x+c_1]{}^2}+a\right )}{b^3 \left (a \text {InverseFunction}\left [\frac {\frac {2 b \text {ArcTan}\left (\frac {\left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right ) a+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\tanh ^{-1}\left (\frac {\text {$\#$1}}{\sqrt {\text {$\#$1}^2+1}}\right )}{a}\&\right ][x+c_1]+i \sqrt {(a-b) (a+b)}\right )}\right )+\log \left (-\frac {2 a^2 \left (i \sqrt {(a-b) (a+b)} \text {InverseFunction}\left [\frac {\frac {2 b \text {ArcTan}\left (\frac {\left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right ) a+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\tanh ^{-1}\left (\frac {\text {$\#$1}}{\sqrt {\text {$\#$1}^2+1}}\right )}{a}\&\right ][x+c_1]+b \sqrt {1+\text {InverseFunction}\left [\frac {\frac {2 b \text {ArcTan}\left (\frac {\left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right ) a+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\tanh ^{-1}\left (\frac {\text {$\#$1}}{\sqrt {\text {$\#$1}^2+1}}\right )}{a}\&\right ][x+c_1]{}^2}+a\right )}{b^3 \left (a \text {InverseFunction}\left [\frac {\frac {2 b \text {ArcTan}\left (\frac {\left (\sqrt {\text {$\#$1}^2+1}-\text {$\#$1}\right ) a+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\tanh ^{-1}\left (\frac {\text {$\#$1}}{\sqrt {\text {$\#$1}^2+1}}\right )}{a}\&\right ][x+c_1]-i \sqrt {(a-b) (a+b)}\right )}\right )\right )}{2 a^2} \\ \end{align*}