Internal problem ID [6857]
Internal file name [OUTPUT/6104_Friday_July_29_2022_03_09_36_AM_52345976/index.tex]
Book: Elementary differential equations. By Earl D. Rainville, Phillip E. Bedient. Macmilliam
Publishing Co. NY. 6th edition. 1981.
Section: CHAPTER 16. Nonlinear equations. Section 101. Independent variable missing.
EXERCISES Page 324
Problem number: 40.
ODE order: 2.
ODE degree: 2.
The type(s) of ODE detected by this program : "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_y]]
\[ \boxed {{y^{\prime \prime }}^{2}-x y^{\prime \prime }+y^{\prime }=0} \]
This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}
Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}
Hence the ode becomes \begin {align*} \left (p^{\prime }\left (x \right )-x \right ) p^{\prime }\left (x \right )+p \left (x \right ) = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. This is Clairaut ODE. It has the form \[ p=p^{\prime }\left (x \right ) x+g\left (p^{\prime }\left (x \right )\right ) \] Where \(g\) is function of \(p'(x)\). Let \(p=p^{\prime }\left (x \right )\) the ode becomes \begin {align*} \left (p -x \right ) p +p = 0 \end {align*}
Solving for \(p \left (x \right )\) from the above results in \begin {align*} p \left (x \right ) &= -\left (p -x \right ) p\tag {1A} \end {align*}
The above ode is a Clairaut ode which is now solved. We start by replacing \(p^{\prime }\left (x \right )\) by \(p\) which gives \begin {align*} p \left (x \right )&=-p^{2}+p x\\ &=-p^{2}+p x \end {align*}
Writing the ode as \begin {align*} p \left (x \right )&= p x +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 \(x\). Hence the above becomes \begin {align*} p = p x +g\tag {1} \end {align*}
Then we see that \begin {align*} g&=-p^{2} \end {align*}
Taking derivative of (1) w.r.t. \(x\) gives \begin {align*} p &=\frac {d}{dx}\left (x p+g\right ) \\ p & =\left ( p+x\frac {dp}{dx}\right ) +\left ( g' \frac {dp}{dx}\right ) \\ p & =p+\left ( x+g'\right ) \frac {dp}{dx}\\ 0 & =\left ( x+g'\right ) \frac {dp}{dx} \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}{dx} & =0\\ p &=c_{1} \end {align*}
Substituting this in (1) gives the general solution as \begin {align*} p \left (x \right ) = -c_{1}^{2}+c_{1} x \end {align*}
The singular solution is found from solving for \(p\) from \begin {align*} x+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*} x+g'\left ( p\right ) &= x -2 p\\ &= 0 \end {align*}
Solving the above for \(p\) results in \begin {align*} p_{1} &=\frac {x}{2} \end {align*}
Substituting the above back in (1) results in \begin {align*} p \left (x \right )_{1} &=\frac {x^{2}}{4} \end {align*}
For solution (1) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = -c_{1}^{2}+c_{1} x \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -c_{1}^{2}+c_{1} x\,\mathop {\mathrm {d}x}}\\ &= \frac {c_{1} x \left (x -2 c_{1} \right )}{2}+c_{2} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \frac {x^{2}}{4} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { \frac {x^{2}}{4}\,\mathop {\mathrm {d}x}}\\ &= \frac {x^{3}}{12}+c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} x \left (x -2 c_{1} \right )}{2}+c_{2} \\ \tag{2} y &= \frac {x^{3}}{12}+c_{3} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} x \left (x -2 c_{1} \right )}{2}+c_{2} \] Verified OK.
\[ y = \frac {x^{3}}{12}+c_{3} \] 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 <- 2nd order ODE linearizable_by_differentiation successful ------------------- * Tackling next 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 <- 2nd order ODE linearizable_by_differentiation successful -> Calling odsolve with the ODE`, diff(y(x), x) = (1/4)*x^2, y(x), singsol = none` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature <- quadrature successful`
✓ Solution by Maple
Time used: 0.219 (sec). Leaf size: 28
dsolve(diff(y(x),x$2)^2-x*diff(y(x),x$2)+diff(y(x),x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {x^{3}}{12}+c_{1} \\ y \left (x \right ) &= \frac {1}{2} c_{1} x^{2}-c_{1}^{2} x +c_{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.002 (sec). Leaf size: 24
DSolve[(y''[x])^2-x*y''[x]+y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {c_1 x^2}{2}-c_1{}^2 x+c_2 \]