Show IVP solution exactness

Question

How do I show that solution of $ y' = sin(ty)$, $y(0)=0$ is exact in $0\leq t\leq \frac{1}{2}$?

I tried to substitute $ty$ with a new variable, but it doesn't lead anywhere.

Answer 1

By Picard-Lindelöf, the IVP

$y'(t)= \sin (ty(t)), \quad y(0)=0$

has a unique solution.

$y(t)=0$ is a solution of the IVP.

Conclusion ?

Answer 2

Let Observe that $$ |y'(t)|=|\sin (ty(t))|\le t|y(t)| $$ for all $t\ge 0$. Set $$ z(t)=\int_0^t |y'(s)|\,ds\ge 0, $$ and thus $$ z'(t)=|y'(t)|\le t|y(t)|=t\left|\int_0^t y'(s)\,ds\,\right|\le t\int_0^t|y'(s)|\,ds=tz(t). $$ Hence $z'\le tz$ and thus $$ \big(\mathrm{e}^{-t^2/2}z(t)\big)'=\mathrm{e}^{-t^2/2}\big(z'(t)-tz(t)\big)\le 0, $$ and integrating in $[0,t]$ we obtain $$ \mathrm{e}^{-t^2/2}z(t)-\mathrm{e}^{0}z(0)\le 0 $$ or $z(t)\le 0$, for all $t\ge 0$. Thus $z'(t)=|y'(t)|\equiv 0$, and hence $y\equiv 0$.