#### Answer

$$ y= \ln(\frac{1}{2}t^2+\frac{1}{2}).$$

#### Work Step by Step

By separation of variables, we have
$$e^{y}dy=tdt $$
then by integration, we get
$$e^y=\frac{1}{2}t^2+c\Longrightarrow y= \ln(\frac{1}{2}t^2+c).$$
Now, since $y(1)=0$, then $c=\frac{1}{2}$.
So the general solution is given by $$ y= \ln(\frac{1}{2}t^2+\frac{1}{2}).$$