Let $y(t)' \leq 2ty(t)+\sqrt{y(t)}$ s.t. $y(0)=0$ where $y(t):[0,\infty) \to [0,\infty)$ continuous and differentiable on $(0,\infty)$. Prove the following $$y(t)\leq \frac{1}{4}\left(e^{t^2/2} \int_{0}^{t}e^{-s^2/2}ds \right)^2$$ for $t\geq0$.
Try: I have proved that one solution for the ode $y(t)' = 2ty(t)+\sqrt{y(t)}$ is exactly $y(t)=\frac{1}{4}\left(e^{t^2/2} \int_{0}^{t}e^{-s^2/2}ds \right)^2$ by differentiating. Now i would like to use a theorem that compares solutions but i need $f(t,y)=2ty+\sqrt{y}$ to be locally Lipschitz for the second variable which is not at zero.And this is where im stuck.
The function $$ f(t) = e^{-t^2} y(t) $$ satisfies $$ f'(t) = e^{-t^2} \left( y'(t) - 2t y(t) \right) \le e^{-t^2} \sqrt{y(t)} = e^{-t^2/2} \sqrt{f(t)} \, . $$ For $t > 0$ with $f(t) > 0$ let $(a, t)$ be the largest interval on which $y$ is not zero, then $f(a) = 0$ and $$ \sqrt{f(t)} = \frac 12 \int_a^t \frac{f'(s)}{\sqrt{f(s)}} \, ds \le \frac 12 \int_0^t e^{-s^2/2} \, ds $$ which implies that $y(t) = e^{t^2} f(t)$ satisfies $$ y(t) \le \frac 14 e^{t^2} \left( \int_0^t e^{-s^2/2} \, ds\right)^2 \, . $$