Internal problem ID [15444]
Internal file name [OUTPUT/15445_Wednesday_May_08_2024_03_58_59_PM_64750972/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 2 (Higher order ODE’s). Section 16. The method of isoclines for differential
equations of the second order. Exercises page 158
Problem number: 701.
ODE order: 2.
ODE degree: 0.
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_x_y1]]
\[ \boxed {x^{\prime \prime }-x \,{\mathrm e}^{x^{\prime }}=0} \]
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(x\) an independent variable. Using \begin {align*} x' &= p(x) \end {align*}
Then \begin {align*} x'' &= \frac {dp}{dt}\\ &= \frac {dx}{dt} \frac {dp}{dx}\\ &= p \frac {dp}{dx} \end {align*}
Hence the ode becomes \begin {align*} p \left (x \right ) \left (\frac {d}{d x}p \left (x \right )\right )-x \,{\mathrm e}^{p \left (x \right )} = 0 \end {align*}
Which is now solved as first order ode for \(p(x)\). In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= \frac {x \,{\mathrm e}^{p}}{p} \end {align*}
Where \(f(x)=x\) and \(g(p)=\frac {{\mathrm e}^{p}}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {{\mathrm e}^{p}}{p}} \,dp &= x \,d x \\ \int { \frac {1}{\frac {{\mathrm e}^{p}}{p}} \,dp} &= \int {x \,d x} \\ -\left (p +1\right ) {\mathrm e}^{-p}&=\frac {x^{2}}{2}+c_{1} \\ \end{align*} The solution is \[ -\left (p \left (x \right )+1\right ) {\mathrm e}^{-p \left (x \right )}-\frac {x^{2}}{2}-c_{1} = 0 \] For solution (1) found earlier, since \(p=x^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} -\left (x^{\prime }+1\right ) {\mathrm e}^{-x^{\prime }}-\frac {x^{2}}{2}-c_{1} = 0 \end {align*}
Integrating both sides gives \begin {align*} \int _{}^{x}\frac {1}{-\operatorname {LambertW}\left (\frac {\left (\textit {\_a}^{2}+2 c_{1} \right ) {\mathrm e}^{-1}}{2}\right )-1}d \textit {\_a} = t +c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{x}\frac {1}{-\operatorname {LambertW}\left (\frac {\left (\textit {\_a}^{2}+2 c_{1} \right ) {\mathrm e}^{-1}}{2}\right )-1}d \textit {\_a} &= t +c_{2} \\ \end{align*}
Verification of solutions
\[ \int _{}^{x}\frac {1}{-\operatorname {LambertW}\left (\frac {\left (\textit {\_a}^{2}+2 c_{1} \right ) {\mathrm e}^{-1}}{2}\right )-1}d \textit {\_a} = t +c_{2} \] 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 differential order: 2; missing variables `, `-> Computing symmetries using: way = 3 `, `-> Computing symmetries using: way = exp_sym -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-_a*exp(_b(_a)) = 0, _b(_a)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable <- separable successful <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 33
dsolve(diff(x(t),t$2)-x(t)*exp(diff(x(t),t))=0,x(t), singsol=all)
\[ -\left (\int _{}^{x \left (t \right )}\frac {1}{\operatorname {LambertW}\left (\frac {\left (\textit {\_a}^{2}+2 c_{1} \right ) {\mathrm e}^{-1}}{2}\right )+1}d \textit {\_a} \right )-t -c_{2} = 0 \]
✓ Solution by Mathematica
Time used: 0.389 (sec). Leaf size: 126
DSolve[x''[t]-x[t]*Exp[x'[t]]==0,x[t],t,IncludeSingularSolutions -> True]
\begin{align*} x(t)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{-W\left (\frac {K[1]^2+2 c_1}{2 e}\right )-1}dK[1]\&\right ][t+c_2] \\ x(t)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{-W\left (\frac {K[1]^2+2 (-1) c_1}{2 e}\right )-1}dK[1]\&\right ][t+c_2] \\ x(t)\to \text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{-W\left (\frac {K[1]^2+2 c_1}{2 e}\right )-1}dK[1]\&\right ][t+c_2] \\ \end{align*}