Internal problem ID [10029]
Internal file name [OUTPUT/8976_Monday_June_06_2022_06_06_42_AM_29441837/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1708 (book 6.117).
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_xy]]
\[ \boxed {y^{\prime \prime } y-{y^{\prime }}^{2}+a y^{\prime } y+y^{2} b=0} \]
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*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) y +\left (-p \left (y \right )+y a \right ) p \left (y \right )+y^{2} b = 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*} u \left (y \right ) y^{2} \left (\left (\frac {d}{d y}u \left (y \right )\right ) y +u \left (y \right )\right )+\left (-u \left (y \right ) y +y a \right ) u \left (y \right ) y = -y^{2} b \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(y,u)\\ &= f( y) g(u)\\ &= -\frac {u a +b}{u y} \end {align*}
Where \(f(y)=-\frac {1}{y}\) and \(g(u)=\frac {u a +b}{u}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u a +b}{u}} \,du &= -\frac {1}{y} \,d y \\ \int { \frac {1}{\frac {u a +b}{u}} \,du} &= \int {-\frac {1}{y} \,d y} \\ -\frac {b \ln \left (u a +b \right )-u a}{a^{2}}&=-\ln \left (y \right )+c_{2} \\ \end{align*} The solution is \[ -\frac {b \ln \left (u \left (y \right ) a +b \right )-u \left (y \right ) a}{a^{2}}+\ln \left (y \right )-c_{2} = 0 \] 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 {b \ln \left (\frac {p \left (y \right ) a}{y}+b \right )-\frac {p \left (y \right ) a}{y}}{a^{2}}+\ln \left (y \right )-c_{2} = 0\\ \frac {-b \ln \left (\frac {p \left (y \right ) a}{y}+b \right )+\frac {p \left (y \right ) a}{y}}{a^{2}}+\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 {-b \ln \left (\frac {y^{\prime } a}{y}+b \right )+\frac {y^{\prime } a}{y}}{a^{2}}+\ln \left (y\right )-c_{2} = 0 \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{\frac {b y \left (-\operatorname {LambertW}\left (-\frac {{\mathrm e}^{\frac {\ln \left (y \right ) a^{2}-c_{2} a^{2}-b}{b}}}{b}\right )+\frac {\ln \left (y \right ) a^{2}-c_{2} a^{2}-b}{b}\right )}{a}-y a \left (\ln \left (y \right )-c_{2} \right )}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\frac {b \textit {\_a} \left (-\operatorname {LambertW}\left (-\frac {{\mathrm e}^{\frac {\ln \left (\textit {\_a} \right ) a^{2}-c_{2} a^{2}-b}{b}}}{b}\right )+\frac {\ln \left (\textit {\_a} \right ) a^{2}-c_{2} a^{2}-b}{b}\right )}{a}-\textit {\_a} a \left (\ln \left (\textit {\_a} \right )-c_{2} \right )}d \textit {\_a}&= c_{3} +x \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\frac {b \textit {\_a} \left (-\operatorname {LambertW}\left (-\frac {{\mathrm e}^{\frac {\ln \left (\textit {\_a} \right ) a^{2}-c_{2} a^{2}-b}{b}}}{b}\right )+\frac {\ln \left (\textit {\_a} \right ) a^{2}-c_{2} a^{2}-b}{b}\right )}{a}-\textit {\_a} a \left (\ln \left (\textit {\_a} \right )-c_{2} \right )}d \textit {\_a} &= c_{3} +x \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {1}{\frac {b \textit {\_a} \left (-\operatorname {LambertW}\left (-\frac {{\mathrm e}^{\frac {\ln \left (\textit {\_a} \right ) a^{2}-c_{2} a^{2}-b}{b}}}{b}\right )+\frac {\ln \left (\textit {\_a} \right ) a^{2}-c_{2} a^{2}-b}{b}\right )}{a}-\textit {\_a} a \left (\ln \left (\textit {\_a} \right )-c_{2} \right )}d \textit {\_a} = c_{3} +x \] 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 a quadrature checking if the LODE has constant coefficients <- constant coefficients successful <- 2nd order, 2 integrating factors of the form mu(x,y) successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 33
dsolve(diff(diff(y(x),x),x)*y(x)-diff(y(x),x)^2+a*y(x)*diff(y(x),x)+y(x)^2*b=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= {\mathrm e}^{\frac {{\mathrm e}^{-a x} c_{1} a +\left (-b x -c_{2} \right ) a +b}{a^{2}}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.404 (sec). Leaf size: 28
DSolve[b*y[x]^2 + a*y[x]*y'[x] - y'[x]^2 + y[x]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 e^{-\frac {b x+c_1 e^{-a x}}{a}} \]