rectify the phase curve of the ODE

Question

Rectify the phase curves of the equation
$$\frac{d^2x}{dt^2}=x-x^2$$in a neighborhood of the point $x=0$ and $\frac{dx}{dt}=1$
I would appreciate helping solve this problem as well a brief explanation or a link so that I can understand what does this question mean by rectifying.

Answer 1

Close to $x=0$ the $x^2$ term will be small, that is, your equation is close to $\ddot x=x$ with solutions $x(t)=x(0)\cosh(t)+x'(0)\sinh(t)$. To get a better approximation, multiply with $2\dot x$ and integrate once to get $$ \dot x^2-x^2+\tfrac23x^3=R^2. $$ Then you can parametrize $(x,\dot x)$ by $(R,u)$ as $\dot x=R\cosh(u)$, $x\sqrt{1-\frac23x}=R\sinh(u)$. Around $x=0$, $|x|\le 1$, this is uniquely solvable, $R=\sqrt{\dot x(0)^2-x(0)^2+\tfrac23x(0)^3}$ is constant along the solution curves and $u={\rm Ar\,sinh}\left(\dfrac xR\sqrt{1-\frac23x}\right)$. This transformation rectifies the solution curves, transforms them into a bundle of parallel lines.

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One can further use this parametrization to approximate the solutions. The function $h(x)=x\sqrt{1-\frac23x}=x-\frac13x^2+...$ is locally invertible around $x=0$, $x=\phi(h(x))=h+\frac13h^2+...$, so that $x=\phi(R\sinh(u))$ and $$ \dot x=\phi'(R\sinh(u))R\cosh(u)\dot u\implies 1=\phi'(R\sinh(u))\dot u=\dot u+\frac23R\sinh(u)\dot u+.... $$ Now with the first terms of the Taylor series of $\phi$ one can compute an approximation of $u$ via $t+c=u+\frac23R(\cosh(u)-1)+...$ which gives a correction $u(t)=t+c-\frac23R(\cosh(t+c)-1)+...$ to the first approximation by the linearized equation.