## Calculus (3rd Edition)

$$4\sqrt y+2\ln y= e^{t^2}+3 .$$
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 $$4\sqrt y+2\ln y= e^{t^2}+3 .$$