7.243 problem 1834 (book 6.243)

Internal problem ID [10155]
Internal file name [OUTPUT/9102_Monday_June_06_2022_06_35_58_AM_90467876/index.tex]

Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1834 (book 6.243).
ODE order: 2.
ODE degree: 2.

The type(s) of ODE detected by this program : "second_order_ode_missing_x"

Maple gives the following as the ode type

[[_2nd_order, _missing_x]]

\[ \boxed {\left (a^{2} y^{2}-b^{2}\right ) {y^{\prime \prime }}^{2}-2 a^{2} y {y^{\prime }}^{2} y^{\prime \prime }+\left ({y^{\prime }}^{2} a^{2}-1\right ) {y^{\prime }}^{2}=0} \]

7.243.1 Solving as second order ode missing x ode

This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using \begin {align*} y' &= p(y) \end {align*}

Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}

Hence the ode becomes \begin {align*} \left (a^{2} y^{2} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-2 a^{2} y p \left (y \right )^{2}-b^{2} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )\right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+\left (a^{2} p \left (y \right )^{3}-p \left (y \right )\right ) p \left (y \right ) = 0 \end {align*}

Which is now solved as first order ode for \(p(y)\). The ode \begin {align*} \left (a^{2} y^{2} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-2 a^{2} y p \left (y \right )^{2}-b^{2} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )\right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+\left (a^{2} p \left (y \right )^{3}-p \left (y \right )\right ) p \left (y \right ) = 0 \end {align*}

is factored to \begin {align*} p \left (y \right )^{2} \left (\left (\frac {d}{d y}p \left (y \right )\right )^{2} a^{2} y^{2}-2 p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) a^{2} y +p \left (y \right )^{2} a^{2}-\left (\frac {d}{d y}p \left (y \right )\right )^{2} b^{2}-1\right ) = 0 \end {align*}

Which gives the following equations \begin {align*} p \left (y \right )^{2} = 0\tag {1} \\ \left (\frac {d}{d y}p \left (y \right )\right )^{2} a^{2} y^{2}-2 p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) a^{2} y +p \left (y \right )^{2} a^{2}-\left (\frac {d}{d y}p \left (y \right )\right )^{2} b^{2}-1 = 0\tag {2} \\ \end {align*}

Each of the above equations is now solved.

Solving ODE (1) Since \(p \left (y \right )^{2} = 0\), is missing derivative in \(p\) then it is an algebraic equation. Solving for \(p \left (y \right )\). \begin {align*} \end {align*}

Solving ODE (2) This is Clairaut ODE. It has the form \[ p=y \left (\frac {d}{d y}p \left (y \right )\right )+g\left (\frac {d}{d y}p \left (y \right )\right ) \] Where \(g\) is function of \(p'(y)\). Let \(p=\frac {d}{d y}p \left (y \right )\) the ode becomes \begin {align*} p^{2} a^{2} y^{2}-2 p p \,a^{2} y +p^{2} a^{2}-p^{2} b^{2} = 1 \end {align*}

Solving for \(p \left (y \right )\) from the above results in \begin {align*} p \left (y \right ) &= \frac {p y a +\sqrt {p^{2} b^{2}+1}}{a}\tag {1A} \\ p \left (y \right ) &= \frac {p y a -\sqrt {p^{2} b^{2}+1}}{a}\tag {2A} \end {align*}

Each of the above ode’s is a Clairaut ode which is now solved. Solving ode 1A We start by replacing \(\frac {d}{d y}p \left (y \right )\) by \(p\) which gives \begin {align*} p \left (y \right )&=p y +\frac {\sqrt {p^{2} b^{2}+1}}{a}\\ &=p y +\frac {\sqrt {p^{2} b^{2}+1}}{a} \end {align*}

Writing the ode as \begin {align*} p \left (y \right )&= p y +g \left (p \right ) \end {align*}

We now write \(g\equiv g\left ( p\right ) \) to make notation simpler but we should always remember that \(g\) is function of \(p\) which in turn is function of \(y\). Hence the above becomes \begin {align*} p = p y +g\tag {1} \end {align*}

Then we see that \begin {align*} g&=\frac {\sqrt {p^{2} b^{2}+1}}{a} \end {align*}

Taking derivative of (1) w.r.t. \(y\) gives \begin {align*} p &=\frac {d}{dy}\left (y p+g\right ) \\ p & =\left ( p+y\frac {dp}{dy}\right ) +\left ( g' \frac {dp}{dy}\right ) \\ p & =p+\left ( y+g'\right ) \frac {dp}{dy}\\ 0 & =\left ( y+g'\right ) \frac {dp}{dy} \end {align*}

Where \(g'\) is derivative of \(g\left ( p\right ) \) w.r.t. \(p\). The general solution is given by \begin {align*} \frac {dp}{dy} & =0\\ p &=c_{1} \end {align*}

Substituting this in (1) gives the general solution as \begin {align*} p \left (y \right ) = y c_{1} +\frac {\sqrt {b^{2} c_{1}^{2}+1}}{a} \end {align*}

The singular solution is found from solving for \(p\) from \begin {align*} y+g'\left ( p\right ) &=0 \end {align*}

And substituting the result back in (1). Since we found above that \(g=\frac {\sqrt {p^{2} b^{2}+1}}{a}\), then the above equation becomes \begin {align*} y+g'\left ( p\right ) &= y +\frac {b^{2} p}{\sqrt {p^{2} b^{2}+1}\, a}\\ &= 0 \end {align*}

Solving the above for \(p\) results in \begin {align*} p_{1} &=\frac {y a}{\sqrt {-a^{2} y^{2}+b^{2}}\, b}\\ p_{2} &=-\frac {y a}{\sqrt {-a^{2} y^{2}+b^{2}}\, b} \end {align*}

Substituting the above back in (1) results in \begin {align*} p \left (y \right )_{1} &=\frac {a^{2} y^{2}+\sqrt {-\frac {b^{2}}{a^{2} y^{2}-b^{2}}}\, \sqrt {-a^{2} y^{2}+b^{2}}\, b}{\sqrt {-a^{2} y^{2}+b^{2}}\, b a}\\ p \left (y \right )_{2} &=-\frac {a^{2} y^{2}-\sqrt {-\frac {b^{2}}{a^{2} y^{2}-b^{2}}}\, \sqrt {-a^{2} y^{2}+b^{2}}\, b}{\sqrt {-a^{2} y^{2}+b^{2}}\, b a} \end {align*}

Solving ode 2A We start by replacing \(\frac {d}{d y}p \left (y \right )\) by \(p\) which gives \begin {align*} p \left (y \right )&=p y -\frac {\sqrt {p^{2} b^{2}+1}}{a}\\ &=p y -\frac {\sqrt {p^{2} b^{2}+1}}{a} \end {align*}

Writing the ode as \begin {align*} p \left (y \right )&= p y +g \left (p \right ) \end {align*}

We now write \(g\equiv g\left ( p\right ) \) to make notation simpler but we should always remember that \(g\) is function of \(p\) which in turn is function of \(y\). Hence the above becomes \begin {align*} p = p y +g\tag {1} \end {align*}

Then we see that \begin {align*} g&=-\frac {\sqrt {p^{2} b^{2}+1}}{a} \end {align*}

Taking derivative of (1) w.r.t. \(y\) gives \begin {align*} p &=\frac {d}{dy}\left (y p+g\right ) \\ p & =\left ( p+y\frac {dp}{dy}\right ) +\left ( g' \frac {dp}{dy}\right ) \\ p & =p+\left ( y+g'\right ) \frac {dp}{dy}\\ 0 & =\left ( y+g'\right ) \frac {dp}{dy} \end {align*}

Where \(g'\) is derivative of \(g\left ( p\right ) \) w.r.t. \(p\). The general solution is given by \begin {align*} \frac {dp}{dy} & =0\\ p &=c_{1} \end {align*}

Substituting this in (1) gives the general solution as \begin {align*} p \left (y \right ) = c_{2} y -\frac {\sqrt {b^{2} c_{2}^{2}+1}}{a} \end {align*}

The singular solution is found from solving for \(p\) from \begin {align*} y+g'\left ( p\right ) &=0 \end {align*}

And substituting the result back in (1). Since we found above that \(g=-\frac {\sqrt {p^{2} b^{2}+1}}{a}\), then the above equation becomes \begin {align*} y+g'\left ( p\right ) &= y -\frac {b^{2} p}{\sqrt {p^{2} b^{2}+1}\, a}\\ &= 0 \end {align*}

Solving the above for \(p\) results in \begin {align*} p_{1} &=-\frac {y a}{\sqrt {-a^{2} y^{2}+b^{2}}\, b}\\ p_{2} &=\frac {y a}{\sqrt {-a^{2} y^{2}+b^{2}}\, b} \end {align*}

Substituting the above back in (1) results in \begin {align*} p \left (y \right )_{1} &=-\frac {a^{2} y^{2}+\sqrt {-\frac {b^{2}}{a^{2} y^{2}-b^{2}}}\, \sqrt {-a^{2} y^{2}+b^{2}}\, b}{\sqrt {-a^{2} y^{2}+b^{2}}\, b a}\\ p \left (y \right )_{2} &=\frac {a^{2} y^{2}-\sqrt {-\frac {b^{2}}{a^{2} y^{2}-b^{2}}}\, \sqrt {-a^{2} y^{2}+b^{2}}\, b}{\sqrt {-a^{2} y^{2}+b^{2}}\, b a} \end {align*}

For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = y c_{1} +\frac {\sqrt {b^{2} c_{1}^{2}+1}}{a} \end {align*}

Integrating both sides gives \begin {align*} \int \frac {a}{y c_{1} a +\sqrt {b^{2} c_{1}^{2}+1}}d y &= c_{3} +x\\ \frac {\ln \left (y c_{1} a +\sqrt {b^{2} c_{1}^{2}+1}\right )}{c_{1}}&=c_{3} +x \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&=\frac {{\mathrm e}^{c_{1} c_{3} +c_{1} x}-\sqrt {b^{2} c_{1}^{2}+1}}{c_{1} a}\\ &=\frac {c_{3} {\mathrm e}^{c_{1} x}-\sqrt {b^{2} c_{1}^{2}+1}}{c_{1} a} \end {align*}

For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = \frac {a^{2} y^{2}+\sqrt {-\frac {b^{2}}{a^{2} y^{2}-b^{2}}}\, \sqrt {-a^{2} y^{2}+b^{2}}\, b}{\sqrt {-a^{2} y^{2}+b^{2}}\, b a} \end {align*}

Integrating both sides gives \begin {align*} \int \frac {\sqrt {-a^{2} y^{2}+b^{2}}\, b a}{a^{2} y^{2}+\sqrt {-\frac {b^{2}}{a^{2} y^{2}-b^{2}}}\, \sqrt {-a^{2} y^{2}+b^{2}}\, b}d y &= \int {dx}\\ \int _{}^{y}\frac {\sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b a}{a^{2} \textit {\_a}^{2}+\sqrt {-\frac {b^{2}}{a^{2} \textit {\_a}^{2}-b^{2}}}\, \sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b}d \textit {\_a}&= x +c_{4} \end {align*}

For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = -\frac {a^{2} y^{2}-\sqrt {-\frac {b^{2}}{a^{2} y^{2}-b^{2}}}\, \sqrt {-a^{2} y^{2}+b^{2}}\, b}{\sqrt {-a^{2} y^{2}+b^{2}}\, b a} \end {align*}

Integrating both sides gives \begin {align*} \int -\frac {\sqrt {-a^{2} y^{2}+b^{2}}\, b a}{a^{2} y^{2}-\sqrt {-\frac {b^{2}}{a^{2} y^{2}-b^{2}}}\, \sqrt {-a^{2} y^{2}+b^{2}}\, b}d y &= \int {dx}\\ \int _{}^{y}-\frac {\sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b a}{a^{2} \textit {\_a}^{2}-\sqrt {-\frac {b^{2}}{a^{2} \textit {\_a}^{2}-b^{2}}}\, \sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b}d \textit {\_a}&= x +c_{5} \end {align*}

For solution (4) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = c_{2} y-\frac {\sqrt {b^{2} c_{2}^{2}+1}}{a} \end {align*}

Integrating both sides gives \begin {align*} \int \frac {a}{c_{2} y a -\sqrt {b^{2} c_{2}^{2}+1}}d y &= x +c_{6}\\ \frac {\ln \left (-c_{2} y a +\sqrt {b^{2} c_{2}^{2}+1}\right )}{c_{2}}&=x +c_{6} \end {align*}

Solving for \(y\) gives these solutions \begin {align*} y_1&=-\frac {{\mathrm e}^{c_{2} c_{6} +c_{2} x}-\sqrt {b^{2} c_{2}^{2}+1}}{c_{2} a}\\ &=-\frac {c_{6} {\mathrm e}^{c_{2} x}-\sqrt {b^{2} c_{2}^{2}+1}}{c_{2} a} \end {align*}

For solution (5) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = -\frac {a^{2} y^{2}+\sqrt {-\frac {b^{2}}{a^{2} y^{2}-b^{2}}}\, \sqrt {-a^{2} y^{2}+b^{2}}\, b}{\sqrt {-a^{2} y^{2}+b^{2}}\, b a} \end {align*}

Integrating both sides gives \begin {align*} \int -\frac {\sqrt {-a^{2} y^{2}+b^{2}}\, b a}{a^{2} y^{2}+\sqrt {-\frac {b^{2}}{a^{2} y^{2}-b^{2}}}\, \sqrt {-a^{2} y^{2}+b^{2}}\, b}d y &= \int {dx}\\ \int _{}^{y}-\frac {\sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b a}{a^{2} \textit {\_a}^{2}+\sqrt {-\frac {b^{2}}{a^{2} \textit {\_a}^{2}-b^{2}}}\, \sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b}d \textit {\_a}&= x +c_{7} \end {align*}

For solution (6) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = \frac {a^{2} y^{2}-\sqrt {-\frac {b^{2}}{a^{2} y^{2}-b^{2}}}\, \sqrt {-a^{2} y^{2}+b^{2}}\, b}{\sqrt {-a^{2} y^{2}+b^{2}}\, b a} \end {align*}

Integrating both sides gives \begin {align*} \int \frac {\sqrt {-a^{2} y^{2}+b^{2}}\, b a}{a^{2} y^{2}-\sqrt {-\frac {b^{2}}{a^{2} y^{2}-b^{2}}}\, \sqrt {-a^{2} y^{2}+b^{2}}\, b}d y &= \int {dx}\\ \int _{}^{y}\frac {\sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b a}{a^{2} \textit {\_a}^{2}-\sqrt {-\frac {b^{2}}{a^{2} \textit {\_a}^{2}-b^{2}}}\, \sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b}d \textit {\_a}&= x +c_{8} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{3} {\mathrm e}^{c_{1} x}-\sqrt {b^{2} c_{1}^{2}+1}}{c_{1} a} \\ \tag{2} \int _{}^{y}\frac {\sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b a}{a^{2} \textit {\_a}^{2}+\sqrt {-\frac {b^{2}}{a^{2} \textit {\_a}^{2}-b^{2}}}\, \sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b}d \textit {\_a} &= x +c_{4} \\ \tag{3} \int _{}^{y}-\frac {\sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b a}{a^{2} \textit {\_a}^{2}-\sqrt {-\frac {b^{2}}{a^{2} \textit {\_a}^{2}-b^{2}}}\, \sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b}d \textit {\_a} &= x +c_{5} \\ \tag{4} y &= -\frac {c_{6} {\mathrm e}^{c_{2} x}-\sqrt {b^{2} c_{2}^{2}+1}}{c_{2} a} \\ \tag{5} \int _{}^{y}-\frac {\sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b a}{a^{2} \textit {\_a}^{2}+\sqrt {-\frac {b^{2}}{a^{2} \textit {\_a}^{2}-b^{2}}}\, \sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b}d \textit {\_a} &= x +c_{7} \\ \tag{6} \int _{}^{y}\frac {\sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b a}{a^{2} \textit {\_a}^{2}-\sqrt {-\frac {b^{2}}{a^{2} \textit {\_a}^{2}-b^{2}}}\, \sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b}d \textit {\_a} &= x +c_{8} \\ \end{align*}

Verification of solutions

\[ y = \frac {c_{3} {\mathrm e}^{c_{1} x}-\sqrt {b^{2} c_{1}^{2}+1}}{c_{1} a} \] Verified OK.

\[ \int _{}^{y}\frac {\sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b a}{a^{2} \textit {\_a}^{2}+\sqrt {-\frac {b^{2}}{a^{2} \textit {\_a}^{2}-b^{2}}}\, \sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b}d \textit {\_a} = x +c_{4} \] Verified OK.

\[ \int _{}^{y}-\frac {\sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b a}{a^{2} \textit {\_a}^{2}-\sqrt {-\frac {b^{2}}{a^{2} \textit {\_a}^{2}-b^{2}}}\, \sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b}d \textit {\_a} = x +c_{5} \] Verified OK.

\[ y = -\frac {c_{6} {\mathrm e}^{c_{2} x}-\sqrt {b^{2} c_{2}^{2}+1}}{c_{2} a} \] Verified OK.

\[ \int _{}^{y}-\frac {\sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b a}{a^{2} \textit {\_a}^{2}+\sqrt {-\frac {b^{2}}{a^{2} \textit {\_a}^{2}-b^{2}}}\, \sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b}d \textit {\_a} = x +c_{7} \] Verified OK.

\[ \int _{}^{y}\frac {\sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b a}{a^{2} \textit {\_a}^{2}-\sqrt {-\frac {b^{2}}{a^{2} \textit {\_a}^{2}-b^{2}}}\, \sqrt {-a^{2} \textit {\_a}^{2}+b^{2}}\, b}d \textit {\_a} = x +c_{8} \] Verified OK.

Maple trace

`Methods for second order ODEs: 
   *** Sublevel 2 *** 
   Methods for second order ODEs: 
   Successful isolation of d^2y/dx^2: 2 solutions were found. Trying to solve each resulting ODE. 
      *** Sublevel 3 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying 2nd order Liouville 
      trying 2nd order WeierstrassP 
      trying 2nd order JacobiSN 
      differential order: 2; trying a linearization to 3rd order 
      trying 2nd order ODE linearizable_by_differentiation 
      trying 2nd order, 2 integrating factors of the form mu(x,y) 
      trying differential order: 2; missing variables 
      `, `-> Computing symmetries using: way = 3 
      `, `-> Computing symmetries using: way = exp_sym 
      -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-(_b(_a)*_a*a^2+(a^2*b^2*_b(_a)^2+_a^2*a^2-b^2)^(1/2))*_b(_a)/(_a^2 
         Methods for first order ODEs: 
         --- Trying classification methods --- 
         trying homogeneous types: 
         trying exact 
         Looking for potential symmetries 
         trying an equivalence to an Abel ODE 
         trying 1st order ODE linearizable_by_differentiation 
         -> Calling odsolve with the ODE`, diff(diff(_b(_a), _a), _a), _b(_a)`            *** Sublevel 5 *** 
            Methods for second order ODEs: 
            --- Trying classification methods --- 
            trying a quadrature 
            <- quadrature successful 
         <- 1st order ODE linearizable_by_differentiation successful 
      <- differential order: 2; canonical coordinates successful 
      <- differential order 2; missing variables successful 
   ------------------- 
   * Tackling next ODE. 
      *** Sublevel 3 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying 2nd order Liouville 
      trying 2nd order WeierstrassP 
      trying 2nd order JacobiSN 
      differential order: 2; trying a linearization to 3rd order 
      trying 2nd order ODE linearizable_by_differentiation 
      trying 2nd order, 2 integrating factors of the form mu(x,y) 
      trying differential order: 2; missing variables 
      `, `-> Computing symmetries using: way = 3 
      `, `-> Computing symmetries using: way = exp_sym 
      -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-(_b(_a)*_a*a^2-(a^2*b^2*_b(_a)^2+_a^2*a^2-b^2)^(1/2))*_b(_a)/(_a^2 
         Methods for first order ODEs: 
         --- Trying classification methods --- 
         trying homogeneous types: 
         trying exact 
         Looking for potential symmetries 
         trying an equivalence to an Abel ODE 
         trying 1st order ODE linearizable_by_differentiation 
         <- 1st order ODE linearizable_by_differentiation successful 
      <- differential order: 2; canonical coordinates successful 
      <- differential order 2; missing variables successful 
-> Calling odsolve with the ODE`, (diff(y(x), x))^2 = (-a^2*y(x)^2+b^2)/(a^2*b^2), y(x), singsol = none`   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   -> Solving 1st order ODE of high degree, 1st attempt 
   trying 1st order WeierstrassP solution for high degree ODE 
   trying 1st order WeierstrassPPrime solution for high degree ODE 
   trying 1st order JacobiSN solution for high degree ODE 
   trying 1st order ODE linearizable_by_differentiation 
   trying differential order: 1; missing variables 
   <- differential order: 1; missing  x  successful 
-> Calling odsolve with the ODE`, diff(y(x), x) = 0, y(x), singsol = none`   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   <- 1st order linear successful`
 

✓ Solution by Maple

Time used: 0.188 (sec). Leaf size: 124

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

\begin{align*} y \left (x \right ) &= \frac {\operatorname {csgn}\left (\sec \left (\frac {c_{1} -x}{b}\right )\right ) \sin \left (\frac {c_{1} -x}{b}\right ) \operatorname {csgn}\left (a \right ) b}{a} \\ y \left (x \right ) &= -\frac {\operatorname {csgn}\left (\sec \left (\frac {c_{1} -x}{b}\right )\right ) \sin \left (\frac {c_{1} -x}{b}\right ) \operatorname {csgn}\left (a \right ) b}{a} \\ y \left (x \right ) &= -\frac {b}{a} \\ y \left (x \right ) &= \frac {b}{a} \\ y \left (x \right ) &= c_{1} \\ y \left (x \right ) &= \frac {b \left ({\mathrm e}^{\frac {\sqrt {c_{1}^{2} a^{2}-1}\, \left (c_{2} +x \right )}{b}}-c_{1} \right )}{\sqrt {c_{1}^{2} a^{2}-1}} \\ \end{align*}

✗ Solution by Mathematica

Time used: 0.0 (sec). Leaf size: 0

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

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