2.29 problem Problem 40(b)

Internal problem ID [12191]
Internal file name [OUTPUT/10844_Thursday_September_21_2023_05_47_56_AM_69214506/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 40(b).
ODE order: 2.
ODE degree: 0.

The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_missing_y"

Maple gives the following as the ode type

[[_2nd_order, _missing_x]]

\[ \boxed {m x^{\prime \prime }-f \left (x^{\prime }\right )=0} \]

2.29.1 Solving as second order ode missing y ode

This is second order ode with missing dependent variable \(x\). Let \begin {align*} p(t) &= x^{\prime } \end {align*}

Then \begin {align*} p'(t) &= x^{\prime \prime } \end {align*}

Hence the ode becomes \begin {align*} m p^{\prime }\left (t \right )-f \left (p \left (t \right )\right ) = 0 \end {align*}

Which is now solve for \(p(t)\) as first order ode. Integrating both sides gives \begin {align*} \int _{}^{p \left (t \right )}\frac {m}{f \left (\textit {\_a} \right )}d \textit {\_a} = t +c_{1} \end {align*}

For solution (1) found earlier, since \(p=x^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \int _{}^{x^{\prime }}\frac {m}{f \left (\textit {\_a} \right )}d \textit {\_a} = t +c_{1} \end {align*}

Integrating both sides gives \begin {align*} x = \int \operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {m}{f \left (\textit {\_a} \right )}d \textit {\_a} \right )+t +c_{1} \right )d t +c_{2} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} x &= \int \operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {m}{f \left (\textit {\_a} \right )}d \textit {\_a} \right )+t +c_{1} \right )d t +c_{2} \\ \end{align*}

Verification of solutions

\[ x = \int \operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {m}{f \left (\textit {\_a} \right )}d \textit {\_a} \right )+t +c_{1} \right )d t +c_{2} \] Verified OK.

2.29.2 Solving as second order ode missing x ode

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*} m p \left (x \right ) \left (\frac {d}{d x}p \left (x \right )\right ) = f \left (p \left (x \right )\right ) \end {align*}

Which is now solved as first order ode for \(p(x)\). Integrating both sides gives \begin {align*} \int _{}^{p \left (x \right )}\frac {m \textit {\_a}}{f \left (\textit {\_a} \right )}d \textit {\_a} = x +c_{1} \end {align*}

For solution (1) found earlier, since \(p=x^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \int _{}^{x^{\prime }}\frac {m \textit {\_a}}{f \left (\textit {\_a} \right )}d \textit {\_a} = x+c_{1} \end {align*}

Integrating both sides gives \begin {align*} \int \frac {1}{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {m \textit {\_a}}{f \left (\textit {\_a} \right )}d \textit {\_a} \right )+x +c_{1} \right )}d x &= \int {dt}\\ \int _{}^{x}\frac {1}{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {m \textit {\_a}}{f \left (\textit {\_a} \right )}d \textit {\_a} \right )+\textit {\_a} +c_{1} \right )}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 {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {m \textit {\_a}}{f \left (\textit {\_a} \right )}d \textit {\_a} \right )+\textit {\_a} +c_{1} \right )}d \textit {\_a} &= t +c_{2} \\ \end{align*}

Verification of solutions

\[ \int _{}^{x}\frac {1}{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {m \textit {\_a}}{f \left (\textit {\_a} \right )}d \textit {\_a} \right )+\textit {\_a} +c_{1} \right )}d \textit {\_a} = t +c_{2} \] 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 
-> Calling odsolve with the ODE`, diff(_b(_a), _a) = f(_b(_a))/m, _b(_a), HINT = [[1, 0]]`   *** Sublevel 2 *** 
   symmetry methods on request 
`, `1st order, trying reduction of order with given symmetries:`[1, 0]
 

✓ Solution by Maple

Time used: 0.016 (sec). Leaf size: 23

dsolve(m*diff(x(t),t$2)=f(diff(x(t),t)),x(t), singsol=all)
 

\[ x \left (t \right ) = \int \operatorname {RootOf}\left (t -m \left (\int _{}^{\textit {\_Z}}\frac {1}{f \left (\textit {\_f} \right )}d \textit {\_f} \right )+c_{1} \right )d t +c_{2} \]

✓ Solution by Mathematica

Time used: 2.257 (sec). Leaf size: 39

DSolve[m*x''[t]==f[x'[t]],x[t],t,IncludeSingularSolutions -> True]
 

\[ x(t)\to \int _1^t\text {InverseFunction}\left [\int _1^{\text {$\#$1}}\frac {1}{f(K[1])}dK[1]\&\right ]\left [c_1+\frac {K[2]}{m}\right ]dK[2]+c_2 \]