Internal problem ID [10066]
Internal file name [OUTPUT/9013_Monday_June_06_2022_06_13_00_AM_87133544/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1745 (book 6.154).
ODE order: 2.
ODE degree: 1.
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], [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {2 y^{\prime \prime } y-{y^{\prime }}^{2} \left ({y^{\prime }}^{2}+1\right )=0} \]
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*} 2 p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) y +\left (-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 \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= \frac {p \left (p^{2}+1\right )}{2 y} \end {align*}
Where \(f(y)=\frac {1}{2 y}\) and \(g(p)=p \left (p^{2}+1\right )\). Integrating both sides gives \begin{align*} \frac {1}{p \left (p^{2}+1\right )} \,dp &= \frac {1}{2 y} \,d y \\ \int { \frac {1}{p \left (p^{2}+1\right )} \,dp} &= \int {\frac {1}{2 y} \,d y} \\ -\frac {\ln \left (p^{2}+1\right )}{2}+\ln \left (p \right )&=\frac {\ln \left (y \right )}{2}+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{-\frac {\ln \left (p^{2}+1\right )}{2}+\ln \left (p \right )} &= {\mathrm e}^{\frac {\ln \left (y \right )}{2}+c_{1}} \end {align*}
Which simplifies to \begin {align*} \frac {p}{\sqrt {p^{2}+1}} &= c_{2} \sqrt {y} \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 } = c_{2} \sqrt {y}\, \sqrt {-\frac {1}{y c_{2}^{2}-1}} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{c_{2} \sqrt {y}\, \sqrt {-\frac {1}{c_{2}^{2} y -1}}}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{c_{2} \sqrt {\textit {\_a}}\, \sqrt {-\frac {1}{\textit {\_a} \,c_{2}^{2}-1}}}d \textit {\_a}&= c_{3} +x \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{c_{2} \sqrt {\textit {\_a}}\, \sqrt {-\frac {1}{\textit {\_a} \,c_{2}^{2}-1}}}d \textit {\_a} &= c_{3} +x \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {1}{c_{2} \sqrt {\textit {\_a}}\, \sqrt {-\frac {1}{\textit {\_a} \,c_{2}^{2}-1}}}d \textit {\_a} = c_{3} +x \] Verified OK.
Maple trace
`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 -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-(1/2)*_b(_a)^2*(_b(_a)^2+1)/_a = 0, _b(_a), HINT = [[_a, 0]]` *** Subl symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 0]
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 331
dsolve(2*diff(diff(y(x),x),x)*y(x)-diff(y(x),x)^2*(diff(y(x),x)^2+1)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= \frac {\left (\operatorname {RootOf}\left (\left (-\cos \left (\textit {\_Z} \right ) c_{1} +c_{1} \textit {\_Z} -2 c_{2} -2 x \right ) \left (\cos \left (\textit {\_Z} \right ) c_{1} +c_{1} \textit {\_Z} -2 c_{2} -2 x \right )\right ) c_{1} -2 x -2 c_{2} \right ) \tan \left (\operatorname {RootOf}\left (\left (-\cos \left (\textit {\_Z} \right ) c_{1} +c_{1} \textit {\_Z} -2 c_{2} -2 x \right ) \left (\cos \left (\textit {\_Z} \right ) c_{1} +c_{1} \textit {\_Z} -2 c_{2} -2 x \right )\right )\right )}{2}+\frac {c_{1}}{2} \\ y \left (x \right ) &= \frac {\left (\operatorname {RootOf}\left (\left (\cos \left (\textit {\_Z} \right ) c_{1} +c_{1} \textit {\_Z} +2 c_{2} +2 x \right ) \left (-\cos \left (\textit {\_Z} \right ) c_{1} +c_{1} \textit {\_Z} +2 c_{2} +2 x \right )\right ) c_{1} +2 x +2 c_{2} \right ) \tan \left (\operatorname {RootOf}\left (\left (\cos \left (\textit {\_Z} \right ) c_{1} +c_{1} \textit {\_Z} +2 c_{2} +2 x \right ) \left (-\cos \left (\textit {\_Z} \right ) c_{1} +c_{1} \textit {\_Z} +2 c_{2} +2 x \right )\right )\right )}{2}+\frac {c_{1}}{2} \\ y \left (x \right ) &= \frac {\left (-\operatorname {RootOf}\left (\left (-\cos \left (\textit {\_Z} \right ) c_{1} +c_{1} \textit {\_Z} -2 c_{2} -2 x \right ) \left (\cos \left (\textit {\_Z} \right ) c_{1} +c_{1} \textit {\_Z} -2 c_{2} -2 x \right )\right ) c_{1} +2 x +2 c_{2} \right ) \tan \left (\operatorname {RootOf}\left (\left (-\cos \left (\textit {\_Z} \right ) c_{1} +c_{1} \textit {\_Z} -2 c_{2} -2 x \right ) \left (\cos \left (\textit {\_Z} \right ) c_{1} +c_{1} \textit {\_Z} -2 c_{2} -2 x \right )\right )\right )}{2}+\frac {c_{1}}{2} \\ y \left (x \right ) &= \frac {\left (-\operatorname {RootOf}\left (\left (\cos \left (\textit {\_Z} \right ) c_{1} +c_{1} \textit {\_Z} +2 c_{2} +2 x \right ) \left (-\cos \left (\textit {\_Z} \right ) c_{1} +c_{1} \textit {\_Z} +2 c_{2} +2 x \right )\right ) c_{1} -2 x -2 c_{2} \right ) \tan \left (\operatorname {RootOf}\left (\left (\cos \left (\textit {\_Z} \right ) c_{1} +c_{1} \textit {\_Z} +2 c_{2} +2 x \right ) \left (-\cos \left (\textit {\_Z} \right ) c_{1} +c_{1} \textit {\_Z} +2 c_{2} +2 x \right )\right )\right )}{2}+\frac {c_{1}}{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 2.67 (sec). Leaf size: 501
DSolve[-(y'[x]^2*(1 + y'[x]^2)) + 2*y[x]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}\left [-i e^{-c_1} \left (\sqrt {\text {$\#$1}} \sqrt {-1+\text {$\#$1} e^{2 c_1}}-e^{-c_1} \text {arctanh}\left (\frac {e^{-c_1} \sqrt {-1+\text {$\#$1} e^{2 c_1}}}{\sqrt {\text {$\#$1}}}\right )\right )\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [i e^{-c_1} \left (\sqrt {\text {$\#$1}} \sqrt {-1+\text {$\#$1} e^{2 c_1}}-e^{-c_1} \text {arctanh}\left (\frac {e^{-c_1} \sqrt {-1+\text {$\#$1} e^{2 c_1}}}{\sqrt {\text {$\#$1}}}\right )\right )\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-i e^{-(-c_1)} \left (\sqrt {\text {$\#$1}} \sqrt {-1+\text {$\#$1} e^{2 (-c_1)}}-e^{-(-c_1)} \text {arctanh}\left (\frac {e^{-(-c_1)} \sqrt {-1+\text {$\#$1} e^{2 (-c_1)}}}{\sqrt {\text {$\#$1}}}\right )\right )\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [i e^{-(-c_1)} \left (\sqrt {\text {$\#$1}} \sqrt {-1+\text {$\#$1} e^{2 (-c_1)}}-e^{-(-c_1)} \text {arctanh}\left (\frac {e^{-(-c_1)} \sqrt {-1+\text {$\#$1} e^{2 (-c_1)}}}{\sqrt {\text {$\#$1}}}\right )\right )\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-i e^{-c_1} \left (\sqrt {\text {$\#$1}} \sqrt {-1+\text {$\#$1} e^{2 c_1}}-e^{-c_1} \text {arctanh}\left (\frac {e^{-c_1} \sqrt {-1+\text {$\#$1} e^{2 c_1}}}{\sqrt {\text {$\#$1}}}\right )\right )\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [i e^{-c_1} \left (\sqrt {\text {$\#$1}} \sqrt {-1+\text {$\#$1} e^{2 c_1}}-e^{-c_1} \text {arctanh}\left (\frac {e^{-c_1} \sqrt {-1+\text {$\#$1} e^{2 c_1}}}{\sqrt {\text {$\#$1}}}\right )\right )\&\right ][x+c_2] \\ \end{align*}