Internal problem ID [10036]
Internal file name [OUTPUT/8983_Monday_June_06_2022_06_07_40_AM_64934202/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1715 (book 6.124).
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]]
\[ \boxed {y^{\prime \prime } y-3 {y^{\prime }}^{2}+3 y y^{\prime }-y^{2}=0} \]
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 (-3 p \left (y \right )+3 y \right ) p \left (y \right )-y^{2} = 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 (-3 u \left (y \right ) y +3 y \right ) u \left (y \right ) y = y^{2} \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(y,u)\\ &= f( y) g(u)\\ &= \frac {2 u^{2}-3 u +1}{u y} \end {align*}
Where \(f(y)=\frac {1}{y}\) and \(g(u)=\frac {2 u^{2}-3 u +1}{u}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {2 u^{2}-3 u +1}{u}} \,du &= \frac {1}{y} \,d y \\ \int { \frac {1}{\frac {2 u^{2}-3 u +1}{u}} \,du} &= \int {\frac {1}{y} \,d y} \\ \ln \left (u -1\right )-\frac {\ln \left (u -\frac {1}{2}\right )}{2}&=\ln \left (y \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{\ln \left (u -1\right )-\frac {\ln \left (u -\frac {1}{2}\right )}{2}} &= {\mathrm e}^{\ln \left (y \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} \frac {\left (u -1\right ) \sqrt {2}}{\sqrt {2 u -1}} &= c_{3} y \end {align*}
The solution is \[ \frac {\left (u \left (y \right )-1\right ) \sqrt {2}}{\sqrt {2 u \left (y \right )-1}} = c_{3} y \] 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 {\left (\frac {p \left (y \right )}{y}-1\right ) \sqrt {2}}{\sqrt {\frac {2 p \left (y \right )}{y}-1}} = c_{3} y\\ \frac {\left (p \left (y \right )-y \right ) \sqrt {2}}{y \sqrt {\frac {2 p \left (y \right )-y}{y}}} = c_{3} 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*} \frac {\left (y^{\prime }-y\right ) \sqrt {2}}{y \sqrt {\frac {2 y^{\prime }-y}{y}}} = c_{3} y \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 }&=\frac {y^{2} c_{3} \sqrt {2}\, \left (\frac {y \sqrt {2}\, c_{3}}{2}+\frac {\sqrt {2 y^{2} c_{3}^{2}+4}}{2}\right )}{2}+y \tag {1} \\ y^{\prime }&=\frac {y^{2} c_{3} \sqrt {2}\, \left (\frac {y \sqrt {2}\, c_{3}}{2}-\frac {\sqrt {2 y^{2} c_{3}^{2}+4}}{2}\right )}{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}{\frac {y^{3} c_{3}^{2}}{2}+\frac {y^{2} c_{3} \sqrt {2}\, \sqrt {2 y^{2} c_{3}^{2}+4}}{4}+y}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\frac {\textit {\_a}^{3} c_{3}^{2}}{2}+\frac {\textit {\_a}^{2} c_{3} \sqrt {2}\, \sqrt {2 c_{3}^{2} \textit {\_a}^{2}+4}}{4}+\textit {\_a}}d \textit {\_a}&= x +c_{4} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int \frac {1}{\frac {y^{3} c_{3}^{2}}{2}-\frac {y^{2} c_{3} \sqrt {2}\, \sqrt {2 y^{2} c_{3}^{2}+4}}{4}+y}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\frac {\textit {\_a}^{3} c_{3}^{2}}{2}-\frac {\textit {\_a}^{2} c_{3} \sqrt {2}\, \sqrt {2 c_{3}^{2} \textit {\_a}^{2}+4}}{4}+\textit {\_a}}d \textit {\_a}&= x +c_{5} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\frac {\textit {\_a}^{3} c_{3}^{2}}{2}+\frac {\textit {\_a}^{2} c_{3} \sqrt {2}\, \sqrt {2 c_{3}^{2} \textit {\_a}^{2}+4}}{4}+\textit {\_a}}d \textit {\_a} &= x +c_{4} \\ \tag{2} \int _{}^{y}\frac {1}{\frac {\textit {\_a}^{3} c_{3}^{2}}{2}-\frac {\textit {\_a}^{2} c_{3} \sqrt {2}\, \sqrt {2 c_{3}^{2} \textit {\_a}^{2}+4}}{4}+\textit {\_a}}d \textit {\_a} &= x +c_{5} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {1}{\frac {\textit {\_a}^{3} c_{3}^{2}}{2}+\frac {\textit {\_a}^{2} c_{3} \sqrt {2}\, \sqrt {2 c_{3}^{2} \textit {\_a}^{2}+4}}{4}+\textit {\_a}}d \textit {\_a} = x +c_{4} \] Verified OK.
\[ \int _{}^{y}\frac {1}{\frac {\textit {\_a}^{3} c_{3}^{2}}{2}-\frac {\textit {\_a}^{2} c_{3} \sqrt {2}\, \sqrt {2 c_{3}^{2} \textit {\_a}^{2}+4}}{4}+\textit {\_a}}d \textit {\_a} = x +c_{5} \] 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: 0.063 (sec). Leaf size: 40
dsolve(diff(diff(y(x),x),x)*y(x)-3*diff(y(x),x)^2+3*y(x)*diff(y(x),x)-y(x)^2=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= \frac {{\mathrm e}^{x}}{\sqrt {2 c_{2} {\mathrm e}^{x}-2 c_{1}}} \\ y \left (x \right ) &= -\frac {{\mathrm e}^{x}}{\sqrt {2 c_{2} {\mathrm e}^{x}-2 c_{1}}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 14.439 (sec). Leaf size: 33
DSolve[-y[x]^2 + 3*y[x]*y'[x] - 3*y'[x]^2 + y[x]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {c_2 e^{x+c_1}}{\sqrt {-1+2 e^{x+c_1}}} \\ y(x)\to 0 \\ \end{align*}