Internal problem ID [10063]
Internal file name [OUTPUT/9010_Monday_June_06_2022_06_12_43_AM_36413461/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1742 (book 6.151).
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]]
\[ \boxed {2 y^{\prime \prime } y-3 {y^{\prime }}^{2}-4 y^{2}=0} \]
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*} 2 p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) y -3 p \left (y \right )^{2}-4 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*} 2 u \left (y \right ) y^{2} \left (\left (\frac {d}{d y}u \left (y \right )\right ) y +u \left (y \right )\right )-3 u \left (y \right )^{2} y^{2} = 4 y^{2} \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(y,u)\\ &= f( y) g(u)\\ &= \frac {u^{2}+4}{2 u y} \end {align*}
Where \(f(y)=\frac {1}{2 y}\) and \(g(u)=\frac {u^{2}+4}{u}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}+4}{u}} \,du &= \frac {1}{2 y} \,d y \\ \int { \frac {1}{\frac {u^{2}+4}{u}} \,du} &= \int {\frac {1}{2 y} \,d y} \\ \frac {\ln \left (u^{2}+4\right )}{2}&=\frac {\ln \left (y \right )}{2}+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {u^{2}+4} &= {\mathrm e}^{\frac {\ln \left (y \right )}{2}+c_{2}} \end {align*}
Which simplifies to \begin {align*} \sqrt {u^{2}+4} &= c_{3} \sqrt {y} \end {align*}
Which simplifies to \[ \sqrt {u \left (y \right )^{2}+4} = c_{3} \sqrt {y}\, {\mathrm e}^{c_{2}} \] The solution is \[ \sqrt {u \left (y \right )^{2}+4} = c_{3} \sqrt {y}\, {\mathrm e}^{c_{2}} \] 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*} \sqrt {\frac {p \left (y \right )^{2}}{y^{2}}+4} = c_{3} \sqrt {y}\, {\mathrm e}^{c_{2}}\\ \sqrt {\frac {p \left (y \right )^{2}+4 y^{2}}{y^{2}}} = c_{3} \sqrt {y}\, {\mathrm e}^{c_{2}} \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*} \sqrt {\frac {{y^{\prime }}^{2}+4 y^{2}}{y^{2}}} = c_{3} \sqrt {y}\, {\mathrm e}^{c_{2}} \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 {-4+c_{3}^{2} {\mathrm e}^{2 c_{2}} y}\, y \tag {1} \\ y^{\prime }&=-\sqrt {-4+c_{3}^{2} {\mathrm e}^{2 c_{2}} y}\, 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 {-4+c_{3}^{2} {\mathrm e}^{2 c_{2}} y}\, y}d y &= \int d x \\ \arctan \left (\frac {\sqrt {-4+c_{3}^{2} {\mathrm e}^{2 c_{2}} y}}{2}\right )&=x +c_{4} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {-4+c_{3}^{2} {\mathrm e}^{2 c_{2}} y}\, y}d y &= \int d x \\ -\arctan \left (\frac {\sqrt {-4+c_{3}^{2} {\mathrm e}^{2 c_{2}} y}}{2}\right )&=x +c_{5} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {4 \left (\tan \left (x +c_{4} \right )^{2}+1\right ) {\mathrm e}^{-2 c_{2}}}{c_{3}^{2}} \\ \tag{2} y &= \frac {4 \left (\tan \left (x +c_{5} \right )^{2}+1\right ) {\mathrm e}^{-2 c_{2}}}{c_{3}^{2}} \\ \end{align*}
Verification of solutions
\[ y = \frac {4 \left (\tan \left (x +c_{4} \right )^{2}+1\right ) {\mathrm e}^{-2 c_{2}}}{c_{3}^{2}} \] Verified OK.
\[ y = \frac {4 \left (\tan \left (x +c_{5} \right )^{2}+1\right ) {\mathrm e}^{-2 c_{2}}}{c_{3}^{2}} \] 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: 0.031 (sec). Leaf size: 22
dsolve(2*diff(diff(y(x),x),x)*y(x)-3*diff(y(x),x)^2-4*y(x)^2=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= \frac {4}{\left (c_{1} \sin \left (x \right )-c_{2} \cos \left (x \right )\right )^{2}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 1.076 (sec). Leaf size: 17
DSolve[-4*y[x]^2 - 3*y'[x]^2 + 2*y[x]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 \sec ^2(x+2 c_1) \]