Solving $\begin{cases}x' = t \sin^2(\frac 1 t) - x^2 \\ x(0) = 0 \end{cases}$ Is there a way to compute the solution of the ode:
$\begin{cases}x' = t \sin^2(\frac 1 t) - x^2 \\ x(0) = 0 \end{cases}$
where in $t = 0$ we define the field as $-x^2$. 
I need for an application of Guiding the solution of ODE with curves. In particular to compute the derivative of  the following at $0$:
$\begin{cases}x' = f(t,x) + \epsilon (\gamma(t) - 2x) \\ x(0) = 0\end{cases}$
 A: Hint:
Let $x=\dfrac{1}{u}\dfrac{du}{dt}$ ,
Then $\dfrac{dx}{dt}=\dfrac{1}{u}\dfrac{d^2u}{dt^2}-\dfrac{1}{u^2}\left(\dfrac{du}{dt}\right)^2$
$\therefore\dfrac{1}{u}\dfrac{d^2u}{dt^2}-\dfrac{1}{u^2}\left(\dfrac{du}{dt}\right)^2=t\sin^2\dfrac{1}{t}-\dfrac{1}{u^2}\left(\dfrac{du}{dt}\right)^2$
$\dfrac{d^2u}{dt^2}-\left(t\sin^2\dfrac{1}{t}\right)u=0$
Let $r=\dfrac{1}{t}$ ,
Then $\dfrac{du}{dt}=\dfrac{du}{dr}\dfrac{dr}{dt}=-\dfrac{1}{t^2}\dfrac{du}{dr}=-r^2\dfrac{du}{dr}$
$\dfrac{d^2u}{dt^2}=\dfrac{d}{dt}\left(-r^2\dfrac{du}{dr}\right)=\dfrac{d}{dr}\left(-r^2\dfrac{du}{dr}\right)\dfrac{dr}{dt}=\left(-r^2\dfrac{d^2u}{dr^2}-2r\dfrac{du}{dr}\right)(-r^2)=r^4\dfrac{d^2u}{dr^2}+2r^3\dfrac{du}{dr}$
$\therefore r^4\dfrac{d^2u}{dr^2}+2r^3\dfrac{du}{dr}-\dfrac{\sin^2r}{r}u=0$
$r\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}-\dfrac{\sin^2r}{r^4}u=0$
Let $v=ru$ ,
Then $\dfrac{dv}{dr}=r\dfrac{du}{dr}+u$
$\dfrac{d^2v}{dr^2}=r\dfrac{d^2u}{dr^2}+\dfrac{du}{dr}+\dfrac{du}{dr}=r\dfrac{d^2u}{dr^2}+2\dfrac{du}{dr}$
$\therefore\dfrac{d^2v}{dr^2}-\dfrac{\sin^2r}{r^5}v=0$
