Internal problem ID [10025]
Internal file name [OUTPUT/8972_Monday_June_06_2022_06_06_16_AM_98507068/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1704 (book 6.113).
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]]
\[ \boxed {y^{\prime \prime } y-{y^{\prime }}^{2}-y^{2} \ln \left (y\right )=0} \]
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*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) y -p \left (y \right )^{2}-y^{2} \ln \left (y \right ) = 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 )-u \left (y \right )^{2} y^{2} = y^{2} \ln \left (y \right ) \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(y,u)\\ &= f( y) g(u)\\ &= \frac {\ln \left (y \right )}{u y} \end {align*}
Where \(f(y)=\frac {\ln \left (y \right )}{y}\) and \(g(u)=\frac {1}{u}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{u}} \,du &= \frac {\ln \left (y \right )}{y} \,d y \\ \int { \frac {1}{\frac {1}{u}} \,du} &= \int {\frac {\ln \left (y \right )}{y} \,d y} \\ \frac {u^{2}}{2}&=\frac {\ln \left (y \right )^{2}}{2}+c_{2} \\ \end{align*} The solution is \[ \frac {u \left (y \right )^{2}}{2}-\frac {\ln \left (y \right )^{2}}{2}-c_{2} = 0 \] 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 {p \left (y \right )^{2}}{2 y^{2}}-\frac {\ln \left (y \right )^{2}}{2}-c_{2} = 0\\ \frac {p \left (y \right )^{2}}{2 y^{2}}-\frac {\ln \left (y \right )^{2}}{2}-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 {{y^{\prime }}^{2}}{2 y^{2}}-\frac {\ln \left (y\right )^{2}}{2}-c_{2} = 0 \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\sqrt {\ln \left (y\right )^{2}+2 c_{2}}\, y \tag {1} \\ y^{\prime }&=-\sqrt {\ln \left (y\right )^{2}+2 c_{2}}\, y \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int \frac {1}{\sqrt {\ln \left (y \right )^{2}+2 c_{2}}\, y}d y &= \int d x \\ \ln \left (\ln \left (y\right )+\sqrt {\ln \left (y\right )^{2}+2 c_{2}}\right )&=x +c_{3} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {\ln \left (y \right )^{2}+2 c_{2}}\, y}d y &= \int d x \\ -\ln \left (\ln \left (y\right )+\sqrt {\ln \left (y\right )^{2}+2 c_{2}}\right )&=x +c_{4} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{\frac {\left ({\mathrm e}^{2 x +2 c_{3}}-2 c_{2} \right ) {\mathrm e}^{-x -c_{3}}}{2}} \\ \tag{2} y &= {\mathrm e}^{\frac {\left ({\mathrm e}^{-2 x -2 c_{4}}-2 c_{2} \right ) {\mathrm e}^{x +c_{4}}}{2}} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{\frac {\left ({\mathrm e}^{2 x +2 c_{3}}-2 c_{2} \right ) {\mathrm e}^{-x -c_{3}}}{2}} \] Verified OK.
\[ y = {\mathrm e}^{\frac {\left ({\mathrm e}^{-2 x -2 c_{4}}-2 c_{2} \right ) {\mathrm e}^{x +c_{4}}}{2}} \] 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: 0.0 (sec). Leaf size: 18
dsolve(diff(diff(y(x),x),x)*y(x)-diff(y(x),x)^2-y(x)^2*ln(y(x))=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{-\frac {c_{1} {\mathrm e}^{x}}{2}+\frac {c_{2} {\mathrm e}^{-x}}{2}} \]
✓ Solution by Mathematica
Time used: 4.551 (sec). Leaf size: 73
DSolve[-(Log[y[x]]*y[x]^2) - y'[x]^2 + y[x]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \exp \left (-\frac {1}{2} \sqrt {c_1} e^{-x-c_2} \left (-1+e^{2 (x+c_2)}\right )\right ) \\ y(x)\to \exp \left (\frac {1}{2} \sqrt {c_1} e^{-x-c_2} \left (-1+e^{2 (x+c_2)}\right )\right ) \\ \end{align*}