60.7.236 problem 1827 (book 6.236)

Internal problem ID [11825]
Book : Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section : Chapter 6, non-linear second order
Problem number : 1827 (book 6.236)
Date solved : Tuesday, January 28, 2025 at 06:23:51 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

\begin{align*} {y^{\prime \prime }}^{2}-a y-b&=0 \end{align*}

Solution by Maple

Time used: 0.122 (sec). Leaf size: 206

dsolve(diff(diff(y(x),x),x)^2-a*y(x)-b=0,y(x), singsol=all)
 
\begin{align*} y &= -\frac {b}{a} \\ a \sqrt {3}\, \left (\int _{}^{y}\frac {1}{\sqrt {a \left (4 \textit {\_a} \sqrt {\textit {\_a} a +b}\, a +4 \sqrt {\textit {\_a} a +b}\, b -c_{1} \right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ -a \sqrt {3}\, \left (\int _{}^{y}\frac {1}{\sqrt {a \left (4 \textit {\_a} \sqrt {\textit {\_a} a +b}\, a +4 \sqrt {\textit {\_a} a +b}\, b -c_{1} \right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ -a \sqrt {3}\, \left (\int _{}^{y}\frac {1}{\sqrt {-a \left (4 \textit {\_a} \sqrt {\textit {\_a} a +b}\, a +4 \sqrt {\textit {\_a} a +b}\, b -c_{1} \right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ a \sqrt {3}\, \left (\int _{}^{y}\frac {1}{\sqrt {-a \left (4 \textit {\_a} \sqrt {\textit {\_a} a +b}\, a +4 \sqrt {\textit {\_a} a +b}\, b -c_{1} \right )}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ \end{align*}

Solution by Mathematica

Time used: 1.230 (sec). Leaf size: 201

DSolve[-b - a*y[x] + D[y[x],{x,2}]^2 == 0,y[x],x,IncludeSingularSolutions -> True]
 
\begin{align*} \text {Solve}\left [\frac {(a y(x)+b)^2 \left (1-\frac {4 (a y(x)+b)^{3/2}}{3 a c_1}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\frac {4 (b+a y(x))^{3/2}}{3 a c_1}\right ){}^2}{a^2 \left (-\frac {4 (a y(x)+b)^{3/2}}{3 a}+c_1\right )}&=(x+c_2){}^2,y(x)\right ] \\ \text {Solve}\left [\frac {(a y(x)+b)^2 \left (1+\frac {4 (a y(x)+b)^{3/2}}{3 a c_1}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},-\frac {4 (b+a y(x))^{3/2}}{3 a c_1}\right ){}^2}{a^2 \left (\frac {4 (a y(x)+b)^{3/2}}{3 a}+c_1\right )}&=(x+c_2){}^2,y(x)\right ] \\ \end{align*}