Internal problem ID [14674]
Internal file name [OUTPUT/14354_Wednesday_April_03_2024_02_17_32_PM_45526039/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.5, page 175
Problem number: 67.
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]]
\[ \boxed {2 y y^{\prime \prime }+y^{2}-{y^{\prime }}^{2}=0} \]
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}{dt}\\ &= \frac {dy}{dt} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}
Hence the ode becomes \begin {align*} 2 y p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+y^{2}-p \left (y \right )^{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 y^{2} u \left (y \right ) \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} \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(y,u)\\ &= f( y) g(u)\\ &= -\frac {u^{2}+1}{2 u y} \end {align*}
Where \(f(y)=-\frac {1}{2 y}\) and \(g(u)=\frac {u^{2}+1}{u}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}+1}{u}} \,du &= -\frac {1}{2 y} \,d y \\ \int { \frac {1}{\frac {u^{2}+1}{u}} \,du} &= \int {-\frac {1}{2 y} \,d y} \\ \frac {\ln \left (u^{2}+1\right )}{2}&=-\frac {\ln \left (y \right )}{2}+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {u^{2}+1} &= {\mathrm e}^{-\frac {\ln \left (y \right )}{2}+c_{2}} \end {align*}
Which simplifies to \begin {align*} \sqrt {u^{2}+1} &= \frac {c_{3}}{\sqrt {y}} \end {align*}
Which simplifies to \[ \sqrt {u \left (y \right )^{2}+1} = \frac {c_{3} {\mathrm e}^{c_{2}}}{\sqrt {y}} \] The solution is \[ \sqrt {u \left (y \right )^{2}+1} = \frac {c_{3} {\mathrm e}^{c_{2}}}{\sqrt {y}} \] 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*} \sqrt {\frac {p \left (y \right )^{2}}{y^{2}}+1} = \frac {c_{3} {\mathrm e}^{c_{2}}}{\sqrt {y}}\\ \sqrt {\frac {y^{2}+p \left (y \right )^{2}}{y^{2}}} = \frac {c_{3} {\mathrm e}^{c_{2}}}{\sqrt {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*} \sqrt {\frac {y^{2}+{y^{\prime }}^{2}}{y^{2}}} = \frac {c_{3} {\mathrm e}^{c_{2}}}{\sqrt {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 }&=\sqrt {c_{3}^{2} y \,{\mathrm e}^{2 c_{2}}-y^{2}} \tag {1} \\ y^{\prime }&=-\sqrt {c_{3}^{2} y \,{\mathrm e}^{2 c_{2}}-y^{2}} \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 {c_{3}^{2} y \,{\mathrm e}^{2 c_{2}}-y^{2}}}d y &= \int d t \\ \arctan \left (\frac {-\frac {c_{3}^{2} {\mathrm e}^{2 c_{2}}}{2}+y}{\sqrt {c_{3}^{2} y \,{\mathrm e}^{2 c_{2}}-y^{2}}}\right )&=t +c_{4} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {c_{3}^{2} y \,{\mathrm e}^{2 c_{2}}-y^{2}}}d y &= \int d t \\ -\arctan \left (\frac {-\frac {c_{3}^{2} {\mathrm e}^{2 c_{2}}}{2}+y}{\sqrt {c_{3}^{2} y \,{\mathrm e}^{2 c_{2}}-y^{2}}}\right )&=t +c_{5} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{3}^{2} {\mathrm e}^{2 c_{2}}}{2}+\frac {\tan \left (t +c_{4} \right ) \sqrt {\frac {c_{3}^{4} {\mathrm e}^{4 c_{2}}}{\tan \left (t +c_{4} \right )^{2}+1}}}{2} \\ \tag{2} y &= \frac {c_{3}^{2} {\mathrm e}^{2 c_{2}}}{2}-\frac {\tan \left (t +c_{5} \right ) \sqrt {\frac {c_{3}^{4} {\mathrm e}^{4 c_{2}}}{\tan \left (t +c_{5} \right )^{2}+1}}}{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{3}^{2} {\mathrm e}^{2 c_{2}}}{2}+\frac {\tan \left (t +c_{4} \right ) \sqrt {\frac {c_{3}^{4} {\mathrm e}^{4 c_{2}}}{\tan \left (t +c_{4} \right )^{2}+1}}}{2} \] Verified OK.
\[ y = \frac {c_{3}^{2} {\mathrm e}^{2 c_{2}}}{2}-\frac {\tan \left (t +c_{5} \right ) \sqrt {\frac {c_{3}^{4} {\mathrm e}^{4 c_{2}}}{\tan \left (t +c_{5} \right )^{2}+1}}}{2} \] 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 -> Calling odsolve with the ODE`, diff(diff(diff(y(t), t), t), t)+diff(y(t), t), y(t)` *** Sublevel 2 *** Methods for third order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients <- constant coefficients successful <- 2nd order ODE linearizable_by_differentiation successful`
✓ Solution by Maple
Time used: 0.063 (sec). Leaf size: 26
dsolve(2*y(t)*diff(y(t),t$2)+y(t)^2=diff(y(t),t)^2,y(t), singsol=all)
\begin{align*} y \left (t \right ) &= 0 \\ y \left (t \right ) &= \sqrt {c_{1}^{2}+c_{2}^{2}}+c_{1} \sin \left (t \right )+c_{2} \cos \left (t \right ) \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.254 (sec). Leaf size: 32
DSolve[2*y[t]*y''[t]+y[t]^2==y'[t]^2,y[t],t,IncludeSingularSolutions -> True]
\begin{align*} y(t)\to c_2 \cos ^2\left (\frac {1}{2} (t-2 c_1)\right ) \\ y(t)\to c_2 \text {Interval}[\{0,1\}] \\ \end{align*}