Consider $\frac{d\psi}{dt}=cos(\psi)$.
Linearizing $\psi$ about $\frac{\pi}{2}$ by setting $\psi= \frac{\pi}{2}+\phi$ , where $\phi$ is very small, we obtain $\frac{d\psi}{dt}=\frac{d\phi}{dt}=-sin \phi$ . Since $\phi$ is very small, $\frac{d\phi}{dt}=-\phi$. Hence the general solution is $\phi= Ae^{-t}$.
Question:
Using $\phi$ , approximate the time it takes for $\psi$ to change from $\psi=\frac{1}{3}$ to $\psi=1$.
My idea consisted of setting $\psi(t)=\frac{\pi}{2}+Ae^{-t}$ and then getting the $t_1$ such that $\psi=1$ and $t_2$ such that $\psi=\frac{1}{3}$ then subtract $t_1-t_2$.
Would this be correct?
$$\frac{d\psi}{dt}=cos(\psi)$$
Without initial condition :
Let $\psi(t)=\tan^{-1}(2y(t))$
$$\frac{d\psi}{dt}=\frac{2}{4y^2+1}\frac{dy}{dt}\quad\text{and}\quad \cos(\psi)=\cos(\tan^{-1}(2y))=\frac{1}{\sqrt{4y^2+1}}$$ $$\frac{dt}{dy}=\frac{2}{\sqrt{4y^2+1}}$$ $$t=\int \frac{2}{\sqrt{4y^2+1}}dy=\sinh^{-1}(2y)+c$$ $$y=\frac12\sinh(t-c)$$ $$\psi(t)=\tan^{-1}(\sinh(t-c))$$