Answer
$y=\displaystyle \frac{1}{2}t^{2}-\frac{1}{4}e^{2t}+(\frac{1}{2}e^{2}-1)t-(\frac{1}{2}+\frac{1}{4}e^{2})$
Work Step by Step
$\displaystyle \frac{d^{2}y}{dt^{2}}=1-e^{2t}\quad $... integrate both sides
... $\displaystyle \quad \int e^{u}du=e^{u}+C, \qquad $so take $\left[\begin{array}{l}
u=2t,\\
du=2dt
\end{array}\right]$
$... \quad $so $\displaystyle \int e^{2t}dt=\frac{1}{2}\int e^{u}d^{u}=\frac{1}{2}e^{2t}+C$
$\displaystyle \frac{dy}{dt}=t-\frac{1}{2}e^{2t}+C\quad $... apply $y'(1)=0$ (to find C)
$0=(1)-\displaystyle \frac{1}{2}e^{2(1)}+C$
$\displaystyle \frac{1}{2}e^{2}-1=C$
So,
$\displaystyle \frac{dy}{dt}=t-\frac{1}{2}e^{2t}+\frac{1}{2}e^{2}-1\quad $... integrate both sides
$y=\displaystyle \frac{1}{2}t^{2}-\frac{1}{2}[\frac{1}{2}e^{2t}]+(\frac{1}{2}e^{2}-1)t+D$
... use the initial value $y(1)=-1$ (to find D)
$-1=-1=\displaystyle \frac{1}{2}-\frac{1}{4}e^{2}+\frac{1}{2}e^{2}-1+D$
$D=-\displaystyle \frac{1}{2}-\frac{1}{4}e^{2}\qquad $ so
$y=\displaystyle \frac{1}{2}t^{2}-\frac{1}{4}e^{2t}+(\frac{1}{2}e^{2}-1)t-(\frac{1}{2}+\frac{1}{4}e^{2})$
