Internal problem ID [12567]
Internal file name [OUTPUT/11220_Wednesday_October_18_2023_10_01_19_PM_60338816/index.tex]
Book: Nonlinear Ordinary Differential Equations by D.W.Jordna and P.Smith. 4th edition 1999.
Oxford Univ. Press. NY
Section: Chapter 2. Plane autonomous systems and linearization. Problems page 79
Problem number: 2.4 (i).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_can_be_made_integrable"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], _Duffing, [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {x^{\prime \prime }+x-x^{3}=0} \]
Multiplying the ode by \(x^{\prime }\) gives \[ x^{\prime } x^{\prime \prime }+\left (1-x^{2}\right ) x^{\prime } x = 0 \] Integrating the above w.r.t \(t\) gives \begin {align*} \int \left (x^{\prime } x^{\prime \prime }+\left (1-x^{2}\right ) x^{\prime } x\right )d t &= 0 \\ \frac {{x^{\prime }}^{2}}{2}-\frac {\left (1-x^{2}\right )^{2}}{4} = c_2 \end {align*}
Which is now solved for \(x\). Solving the given ode for \(x^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} x^{\prime }&=\frac {\sqrt {2+2 x^{4}-4 x^{2}+8 c_{1}}}{2} \tag {1} \\ x^{\prime }&=-\frac {\sqrt {2+2 x^{4}-4 x^{2}+8 c_{1}}}{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 {2}{\sqrt {2 x^{4}-4 x^{2}+8 c_{1} +2}}d x &= \int {dt}\\ \int _{}^{x}\frac {2}{\sqrt {2 \textit {\_a}^{4}-4 \textit {\_a}^{2}+8 c_{1} +2}}d \textit {\_a}&= t +c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {2}{\sqrt {2 x^{4}-4 x^{2}+8 c_{1} +2}}d x &= \int {dt}\\ \int _{}^{x}-\frac {2}{\sqrt {2 \textit {\_a}^{4}-4 \textit {\_a}^{2}+8 c_{1} +2}}d \textit {\_a}&= t +c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{x}\frac {2}{\sqrt {2 \textit {\_a}^{4}-4 \textit {\_a}^{2}+8 c_{1} +2}}d \textit {\_a} &= t +c_{2} \\ \tag{2} \int _{}^{x}-\frac {2}{\sqrt {2 \textit {\_a}^{4}-4 \textit {\_a}^{2}+8 c_{1} +2}}d \textit {\_a} &= t +c_{3} \\ \end{align*}
Verification of solutions
\[ \int _{}^{x}\frac {2}{\sqrt {2 \textit {\_a}^{4}-4 \textit {\_a}^{2}+8 c_{1} +2}}d \textit {\_a} = t +c_{2} \] Verified OK.
\[ \int _{}^{x}-\frac {2}{\sqrt {2 \textit {\_a}^{4}-4 \textit {\_a}^{2}+8 c_{1} +2}}d \textit {\_a} = t +c_{3} \] Verified OK.
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 )+\left (-x^{2}+1\right ) x = 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 {\left (x^{2}-1\right ) x}{p} \end {align*}
Where \(f(x)=\left (x^{2}-1\right ) x\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= \left (x^{2}-1\right ) x \,d x \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {\left (x^{2}-1\right ) x \,d x} \\ \frac {p^{2}}{2}&=\frac {\left (x^{2}-1\right )^{2}}{4}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (x \right )^{2}}{2}-\frac {\left (x^{2}-1\right )^{2}}{4}-c_{1} = 0 \] For solution (1) found earlier, since \(p=x^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \frac {{x^{\prime }}^{2}}{2}-\frac {\left (x^{2}-1\right )^{2}}{4}-c_{1} = 0 \end {align*}
Solving the given ode for \(x^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} x^{\prime }&=\frac {\sqrt {2+2 x^{4}-4 x^{2}+8 c_{1}}}{2} \tag {1} \\ x^{\prime }&=-\frac {\sqrt {2+2 x^{4}-4 x^{2}+8 c_{1}}}{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 {2}{\sqrt {2 x^{4}-4 x^{2}+8 c_{1} +2}}d x &= \int {dt}\\ \int _{}^{x}\frac {2}{\sqrt {2 \textit {\_a}^{4}-4 \textit {\_a}^{2}+8 c_{1} +2}}d \textit {\_a}&= t +c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {2}{\sqrt {2 x^{4}-4 x^{2}+8 c_{1} +2}}d x &= \int {dt}\\ \int _{}^{x}-\frac {2}{\sqrt {2 \textit {\_a}^{4}-4 \textit {\_a}^{2}+8 c_{1} +2}}d \textit {\_a}&= t +c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{x}\frac {2}{\sqrt {2 \textit {\_a}^{4}-4 \textit {\_a}^{2}+8 c_{1} +2}}d \textit {\_a} &= t +c_{2} \\ \tag{2} \int _{}^{x}-\frac {2}{\sqrt {2 \textit {\_a}^{4}-4 \textit {\_a}^{2}+8 c_{1} +2}}d \textit {\_a} &= t +c_{3} \\ \end{align*}
Verification of solutions
\[ \int _{}^{x}\frac {2}{\sqrt {2 \textit {\_a}^{4}-4 \textit {\_a}^{2}+8 c_{1} +2}}d \textit {\_a} = t +c_{2} \] Verified OK.
\[ \int _{}^{x}-\frac {2}{\sqrt {2 \textit {\_a}^{4}-4 \textit {\_a}^{2}+8 c_{1} +2}}d \textit {\_a} = t +c_{3} \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP trying 2nd order JacobiSN <- 2nd_order JacobiSN successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 43
dsolve(diff(x(t),t$2)+x(t)-x(t)^3=0,x(t), singsol=all)
\[ x \left (t \right ) = c_{2} \sqrt {2}\, \sqrt {\frac {1}{c_{2}^{2}+1}}\, \operatorname {JacobiSN}\left (\frac {\left (\sqrt {2}\, t +2 c_{1} \right ) \sqrt {2}\, \sqrt {\frac {1}{c_{2}^{2}+1}}}{2}, c_{2}\right ) \]
✓ Solution by Mathematica
Time used: 60.266 (sec). Leaf size: 171
DSolve[x''[t]+x[t]-x[t]^3==0,x[t],t,IncludeSingularSolutions -> True]
\begin{align*} x(t)\to -\frac {i \text {sn}\left (\frac {\sqrt {\left (\sqrt {1-2 c_1}+1\right ) (t+c_2){}^2}}{\sqrt {2}}|\frac {1-\sqrt {1-2 c_1}}{\sqrt {1-2 c_1}+1}\right )}{\sqrt {\frac {1}{-1+\sqrt {1-2 c_1}}}} \\ x(t)\to \frac {i \text {sn}\left (\frac {\sqrt {\left (\sqrt {1-2 c_1}+1\right ) (t+c_2){}^2}}{\sqrt {2}}|\frac {1-\sqrt {1-2 c_1}}{\sqrt {1-2 c_1}+1}\right )}{\sqrt {\frac {1}{-1+\sqrt {1-2 c_1}}}} \\ \end{align*}