Internal problem ID [12184]
Internal file name [OUTPUT/10837_Thursday_September_21_2023_05_47_44_AM_26570493/index.tex]
Book: Differential equations and the calculus of variations by L. ElSGOLTS. MIR PUBLISHERS,
MOSCOW, Third printing 1977.
Section: Chapter 2, DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER.
Problems page 172
Problem number: Problem 31.
ODE order: 2.
ODE degree: 2.
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]]
\[ \boxed {y y^{\prime } y^{\prime \prime }-{y^{\prime }}^{3}-{y^{\prime \prime }}^{2}=0} \]
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}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}
Hence the ode becomes \begin {align*} \left (p \left (y \right ) y -p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )\right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-p \left (y \right )^{3} = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). The ode \begin {align*} \left (p \left (y \right ) y -p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )\right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )-p \left (y \right )^{3} = 0 \end {align*}
is factored to \begin {align*} p \left (y \right )^{2} \left (\left (\frac {d}{d y}p \left (y \right )\right )^{2}-\left (\frac {d}{d y}p \left (y \right )\right ) y +p \left (y \right )\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} p \left (y \right )^{2} = 0\tag {1} \\ \left (\frac {d}{d y}p \left (y \right )\right )^{2}-\left (\frac {d}{d y}p \left (y \right )\right ) y +p \left (y \right ) = 0\tag {2} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) Since \(p \left (y \right )^{2} = 0\), is missing derivative in \(p\) then it is an algebraic equation. Solving for \(p \left (y \right )\). \begin {align*} \end {align*}
Solving ODE (2) This is Clairaut ODE. It has the form \[ p=\left (\frac {d}{d y}p \left (y \right )\right ) y+g\left (\frac {d}{d y}p \left (y \right )\right ) \] Where \(g\) is function of \(p'(y)\). Let \(p=\frac {d}{d y}p \left (y \right )\) the ode becomes \begin {align*} p^{2}-p y +p = 0 \end {align*}
Solving for \(p \left (y \right )\) from the above results in \begin {align*} p \left (y \right ) &= -p^{2}+p y\tag {1A} \end {align*}
The above ode is a Clairaut ode which is now solved. We start by replacing \(\frac {d}{d y}p \left (y \right )\) by \(p\) which gives \begin {align*} p \left (y \right )&=-p^{2}+p y\\ &=-p^{2}+p y \end {align*}
Writing the ode as \begin {align*} p \left (y \right )&= p y +g \left (p \right ) \end {align*}
We now write \(g\equiv g\left ( p\right ) \) to make notation simpler but we should always remember that \(g\) is function of \(p\) which in turn is function of \(y\). Hence the above becomes \begin {align*} p = p y +g\tag {1} \end {align*}
Then we see that \begin {align*} g&=-p^{2} \end {align*}
Taking derivative of (1) w.r.t. \(y\) gives \begin {align*} p &=\frac {d}{dy}\left (y p+g\right ) \\ p & =\left ( p+y\frac {dp}{dy}\right ) +\left ( g' \frac {dp}{dy}\right ) \\ p & =p+\left ( y+g'\right ) \frac {dp}{dy}\\ 0 & =\left ( y+g'\right ) \frac {dp}{dy} \end {align*}
Where \(g'\) is derivative of \(g\left ( p\right ) \) w.r.t. \(p\). The general solution is given by \begin {align*} \frac {dp}{dy} & =0\\ p &=c_{1} \end {align*}
Substituting this in (1) gives the general solution as \begin {align*} p \left (y \right ) = -c_{1}^{2}+c_{1} y \end {align*}
The singular solution is found from solving for \(p\) from \begin {align*} y+g'\left ( p\right ) &=0 \end {align*}
And substituting the result back in (1). Since we found above that \(g=-p^{2}\), then the above equation becomes \begin {align*} y+g'\left ( p\right ) &= y -2 p\\ &= 0 \end {align*}
Solving the above for \(p\) results in \begin {align*} p_{1} &=\frac {y}{2} \end {align*}
Substituting the above back in (1) results in \begin {align*} p \left (y \right )_{1} &=\frac {y^{2}}{4} \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*} y^{\prime } = -c_{1}^{2}+c_{1} y \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{-c_{1}^{2}+c_{1} y}d y &= x +c_{2}\\ \frac {\ln \left (-c_{1} +y \right )}{c_{1}}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&={\mathrm e}^{c_{1} c_{2} +c_{1} x}+c_{1}\\ &=c_{2} {\mathrm e}^{c_{1} x}+c_{1} \end {align*}
For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = \frac {y^{2}}{4} \end {align*}
Integrating both sides gives \begin {align*} \int \frac {4}{y^{2}}d y &= x +c_{3}\\ -\frac {4}{y}&=x +c_{3} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&=-\frac {4}{x +c_{3}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{2} {\mathrm e}^{c_{1} x}+c_{1} \\ \tag{2} y &= -\frac {4}{x +c_{3}} \\ \end{align*}
Verification of solutions
\[ y = c_{2} {\mathrm e}^{c_{1} x}+c_{1} \] Verified OK.
\[ y = -\frac {4}{x +c_{3}} \] Verified OK.
Maple trace
`Methods for second order ODEs: *** Sublevel 2 *** Methods for second order ODEs: Successful isolation of d^2y/dx^2: 2 solutions were found. Trying to solve each resulting ODE. *** Sublevel 3 *** 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 differential order: 2; missing variables `, `-> Computing symmetries using: way = 3 -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)+(1/2)*(-_a+(_a^2-4*_b(_a))^(1/2))*_b(_a) = 0, _b(_a), HINT = [[_a, symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 2*_b]
✓ Solution by Maple
Time used: 7.281 (sec). Leaf size: 42
dsolve(y(x)*diff(y(x),x)*diff(y(x),x$2)=diff(y(x),x)^3+diff(y(x),x$2)^2,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -\frac {4}{-4 c_{1} +x} \\ y \left (x \right ) &= c_{1} \\ y \left (x \right ) &= {\mathrm e}^{-c_{1} \left (c_{2} +x \right )}-c_{1} \\ y \left (x \right ) &= {\mathrm e}^{c_{1} \left (c_{2} +x \right )}+c_{1} \\ \end{align*}
✓ Solution by Mathematica
Time used: 13.794 (sec). Leaf size: 119
DSolve[y[x]*y'[x]*y''[x]==y'[x]^3+y''[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {1}{2} \left (e^{-\frac {1}{2} \left (1+e^{c_1}\right ) (x+c_2)}-1-e^{c_1}\right ) \\ y(x)\to \frac {1+e^{\frac {x+c_2}{-1+\tanh \left (\frac {c_1}{2}\right )}}}{-1+\tanh \left (\frac {c_1}{2}\right )} \\ y(x)\to -\frac {1}{2}-\frac {1}{2} e^{-\frac {x}{2}-\frac {c_2}{2}} \\ y(x)\to \frac {1}{2} \left (-1+e^{-\frac {x}{2}-\frac {c_2}{2}}\right ) \\ \end{align*}