Answer
$$ 4\sqrt y+2\ln y= e^{t^2}+3 .$$
Work Step by Step
By separation of variables, we have
$$(y^{-1/2}+y^{-1})dy=te^{t^2} dt $$
then by integration, we get
$$ \frac{1}{1/2}\sqrt y+\ln y=\frac{1}{2}e^{t^2}+c \Longrightarrow4\sqrt y+2\ln y= e^{t^2}+ A .$$
Now, since $y(0)=1$, then $A=3$.
So we have $$ 4\sqrt y+2\ln y= e^{t^2}+3 .$$
