# straighten out the flow of ode

I've been ready G.Teschl ODE book and I got really confused with how to apply this theorem:

(Straighten out of vector fields)

Suppose that $$f(x_{0})\neq 0$$. Then, there is a local coordinate transform $$y=\phi(x)$$ s.t $$\dot x = f(x)$$ is transformed to $$\dot y = (1,0,...,0)$$.

I think I can picture the theorem and I understand the proof, but my real question is how to apply it. For example, there is this problem in the book which seems to be a direct use for this theorem and I have no idea how to solve it.

REAL PROBLEM

Find a transformation which straighten out the flow $$\dot x = x$$, defined in $$M=\mathbb{R}$$

Any help would be really appreciated <3

If $$x\neq 0$$ (this is equivalent to the condition $$f(x_{0})\neq 0$$ in the hypotheses for the "straightening" lemma), then we can write $$\dfrac{x'}{x}=1.$$ We know that the general solution of the ODE is $$x(t)=Ce^t$$ for some constant $$C$$. Thus, $$x(t)$$ and $$x'(t)$$ have the same sign as $$C$$ for all $$t$$, and consequently, $$\frac{d}{dt}\log \left\lvert x(t) \right\rvert = \dfrac{x'(t)}{x(t)}$$ by the chain rule. Then, we can choose $$y=\phi(x)=\log\left\lvert x \right\rvert$$ as our local coordinate transform. We conclude that for all non-zero $$x$$, the equation $$x'=x$$ can be written as $$y'=1$$ under this change of coordinates.
• +1, but you need to be a bit more careful with the case $x<0$. – Hans Lundmark Nov 9 at 10:42
• Good point. One could argue that $\phi(x) = \log \left\lvert x \right\rvert$ works, if we assume that the general solution of the ODE is $x(t)=Ce^t$ and thus $x(t)$ and $x'(t)$ always have the same sign as $C$. I'll edit my answer. – B. Núñez Nov 9 at 18:12