7.232 problem 1823 (book 6.232)

Internal problem ID [10144]
Internal file name [OUTPUT/9091_Monday_June_06_2022_06_29_43_AM_52120896/index.tex]

Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1823 (book 6.232).
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]]

\[ \boxed {\left ({y^{\prime }}^{2}+y^{2}\right ) y^{\prime \prime }+y^{3}=0} \]

7.232.1 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 (p \left (y \right )^{2}+y^{2}\right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+y^{3} = 0 \end {align*}

Which is now solved as first order ode for \(p(y)\). Using the change of variables \(p \left (y \right ) = u \left (y \right ) y\) on the above ode results in new ode in \(u \left (y \right )\) \begin {align*} \left (u \left (y \right )^{2} y^{2}+y^{2}\right ) u \left (y \right ) y \left (\left (\frac {d}{d y}u \left (y \right )\right ) y +u \left (y \right )\right ) = -y^{3} \end {align*}

In canonical form the ODE is \begin {align*} u' &= F(y,u)\\ &= f( y) g(u)\\ &= -\frac {u^{4}+u^{2}+1}{u y \left (u^{2}+1\right )} \end {align*}

Where \(f(y)=-\frac {1}{y}\) and \(g(u)=\frac {u^{4}+u^{2}+1}{u \left (u^{2}+1\right )}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{4}+u^{2}+1}{u \left (u^{2}+1\right )}} \,du &= -\frac {1}{y} \,d y \\ \int { \frac {1}{\frac {u^{4}+u^{2}+1}{u \left (u^{2}+1\right )}} \,du} &= \int {-\frac {1}{y} \,d y} \\ \frac {\ln \left (u^{4}+u^{2}+1\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 u^{2}+1\right ) \sqrt {3}}{3}\right )}{6}&=-\ln \left (y \right )+c_{2} \\ \end{align*} The solution is \[ \frac {\ln \left (u \left (y \right )^{4}+u \left (y \right )^{2}+1\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 u \left (y \right )^{2}+1\right ) \sqrt {3}}{3}\right )}{6}+\ln \left (y \right )-c_{2} = 0 \] Replacing \(u(y)\) in the above solution by \(\frac {p \left (y \right )}{y}\) results in the solution for \(p \left (y \right )\) in implicit form \begin {align*} \frac {\ln \left (\frac {p \left (y \right )^{4}}{y^{4}}+\frac {p \left (y \right )^{2}}{y^{2}}+1\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (\frac {2 p \left (y \right )^{2}}{y^{2}}+1\right ) \sqrt {3}}{3}\right )}{6}+\ln \left (y \right )-c_{2} = 0\\ \frac {\ln \left (\frac {p \left (y \right )^{4}}{y^{4}}+\frac {p \left (y \right )^{2}}{y^{2}}+1\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 p \left (y \right )^{2}+y^{2}\right ) \sqrt {3}}{3 y^{2}}\right )}{6}+\ln \left (y \right )-c_{2} = 0 \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*} \frac {\ln \left (\frac {{y^{\prime }}^{4}}{y^{4}}+\frac {{y^{\prime }}^{2}}{y^{2}}+1\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 {y^{\prime }}^{2}+y^{2}\right ) \sqrt {3}}{3 y^{2}}\right )}{6}+\ln \left (y\right )-c_{2} = 0 \end {align*}

Integrating both sides gives \begin {align*} \int \frac {1}{\operatorname {RootOf}\left (2 \textit {\_Z}^{2}-\sqrt {3}\, \tan \left (\operatorname {RootOf}\left (-4 \sqrt {3}\, \ln \left (y \right )-\sqrt {3}\, \ln \left (\frac {3}{4}+\frac {3 \tan \left (\textit {\_Z} \right )^{2}}{4}\right )+4 \sqrt {3}\, c_{2} -2 \textit {\_Z} \right )\right )+1\right ) y}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\operatorname {RootOf}\left (2 \textit {\_Z}^{2}-\sqrt {3}\, \tan \left (\operatorname {RootOf}\left (-4 \sqrt {3}\, \ln \left (\textit {\_a} \right )-\sqrt {3}\, \ln \left (\frac {3}{4}+\frac {3 \tan \left (\textit {\_Z} \right )^{2}}{4}\right )+4 \sqrt {3}\, c_{2} -2 \textit {\_Z} \right )\right )+1\right ) \textit {\_a}}d \textit {\_a}&= c_{3} +x \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\operatorname {RootOf}\left (2 \textit {\_Z}^{2}-\sqrt {3}\, \tan \left (\operatorname {RootOf}\left (-4 \sqrt {3}\, \ln \left (\textit {\_a} \right )-\sqrt {3}\, \ln \left (\frac {3}{4}+\frac {3 \tan \left (\textit {\_Z} \right )^{2}}{4}\right )+4 \sqrt {3}\, c_{2} -2 \textit {\_Z} \right )\right )+1\right ) \textit {\_a}}d \textit {\_a} &= c_{3} +x \\ \end{align*}

Verification of solutions

\[ \int _{}^{y}\frac {1}{\operatorname {RootOf}\left (2 \textit {\_Z}^{2}-\sqrt {3}\, \tan \left (\operatorname {RootOf}\left (-4 \sqrt {3}\, \ln \left (\textit {\_a} \right )-\sqrt {3}\, \ln \left (\frac {3}{4}+\frac {3 \tan \left (\textit {\_Z} \right )^{2}}{4}\right )+4 \sqrt {3}\, c_{2} -2 \textit {\_Z} \right )\right )+1\right ) \textit {\_a}}d \textit {\_a} = c_{3} +x \] 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 
Try integration with the canonical coordinates of the symmetry [0, y] 
-> Calling odsolve with the ODE`, diff(_b(_a), _a) = -(_b(_a)^4+_b(_a)^2+1)/(_b(_a)^2+1), _b(_a), explicit, HINT = [[1, 0]]`   *** S 
   symmetry methods on request 
`, `1st order, trying reduction of order with given symmetries:`[1, 0]
 

✓ Solution by Maple

Time used: 0.078 (sec). Leaf size: 163

dsolve((diff(y(x),x)^2+y(x)^2)*diff(diff(y(x),x),x)+y(x)^3=0,y(x), singsol=all)
 

\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= \frac {\sqrt {c_{1} +\tan \left (\sqrt {3}\, x \right )}\, {\mathrm e}^{-\frac {\sqrt {3}\, \left (\int \frac {\sqrt {\left (9 c_{1}^{2}+12\right ) \sec \left (\sqrt {3}\, x \right )^{2}+3 c_{1}^{2}+6 c_{1} \tan \left (\sqrt {3}\, x \right )-3}}{c_{1} +\tan \left (\sqrt {3}\, x \right )}d x \right )}{6}+c_{2}}}{\left (\sec \left (\sqrt {3}\, x \right )^{2}\right )^{\frac {1}{4}}} \\ y \left (x \right ) &= \frac {\sqrt {c_{1} +\tan \left (\sqrt {3}\, x \right )}\, {\mathrm e}^{\frac {\sqrt {3}\, \left (\int \frac {\sqrt {\left (9 c_{1}^{2}+12\right ) \sec \left (\sqrt {3}\, x \right )^{2}+3 c_{1}^{2}+6 c_{1} \tan \left (\sqrt {3}\, x \right )-3}}{c_{1} +\tan \left (\sqrt {3}\, x \right )}d x \right )}{6}+c_{2}}}{\left (\sec \left (\sqrt {3}\, x \right )^{2}\right )^{\frac {1}{4}}} \\ \end{align*}

✓ Solution by Mathematica

Time used: 60.991 (sec). Leaf size: 369

DSolve[y[x]^3 + (y[x]^2 + y'[x]^2)*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {c_2 \exp \left (-\frac {\arctan \left (\frac {1+2 \text {InverseFunction}\left [\frac {\left (\sqrt {3}-i\right ) \arctan \left (\frac {\text {$\#$1}}{\sqrt {\frac {1}{2} \left (1-i \sqrt {3}\right )}}\right )}{\sqrt {6 \left (1-i \sqrt {3}\right )}}+\frac {\left (\sqrt {3}+i\right ) \arctan \left (\frac {\text {$\#$1}}{\sqrt {\frac {1}{2} \left (1+i \sqrt {3}\right )}}\right )}{\sqrt {6 \left (1+i \sqrt {3}\right )}}\&\right ][-x+c_1]{}^2}{\sqrt {3}}\right )}{2 \sqrt {3}}\right )}{\sqrt [4]{\text {InverseFunction}\left [\frac {\left (\sqrt {3}-i\right ) \arctan \left (\frac {\text {$\#$1}}{\sqrt {\frac {1}{2} \left (1-i \sqrt {3}\right )}}\right )}{\sqrt {6 \left (1-i \sqrt {3}\right )}}+\frac {\left (\sqrt {3}+i\right ) \arctan \left (\frac {\text {$\#$1}}{\sqrt {\frac {1}{2} \left (1+i \sqrt {3}\right )}}\right )}{\sqrt {6 \left (1+i \sqrt {3}\right )}}\&\right ][-x+c_1]{}^4+\text {InverseFunction}\left [\frac {\left (\sqrt {3}-i\right ) \arctan \left (\frac {\text {$\#$1}}{\sqrt {\frac {1}{2} \left (1-i \sqrt {3}\right )}}\right )}{\sqrt {6 \left (1-i \sqrt {3}\right )}}+\frac {\left (\sqrt {3}+i\right ) \arctan \left (\frac {\text {$\#$1}}{\sqrt {\frac {1}{2} \left (1+i \sqrt {3}\right )}}\right )}{\sqrt {6 \left (1+i \sqrt {3}\right )}}\&\right ][-x+c_1]{}^2+1}} \]