Answer
$y=2e^{-x}+2x-1$
Work Step by Step
$\displaystyle \frac{d^{2}y}{dx^{2}}=2e^{-x}\quad $... integrate both sides
... $\displaystyle \quad \int e^{u}du=e^{u}+C, \qquad $so take $\left[\begin{array}{l}
u=-x,\\
du=-dx
\end{array}\right]$
$... \quad $so $2\displaystyle \int e^{-x}dx=-2\int e^{u}d^{u}=-2e^{-x}+C$
$\displaystyle \frac{dy}{dx}=-2e^{-2x}+D\qquad $... find C using $y'(0)=1$
$1=-2e^{0}+D$
$2=D$
So,
$\displaystyle \frac{dy}{dx}=-2e^{-x}+2\quad $... integrate both sides (use same sub as above)
$... \displaystyle \int 2e^{-x}dx=2e^{-x}+C$
$y=2e^{-x}+2x+D\qquad $... find $D$ using $y(0)=1$
$1=2e^{0}+0+D$
$D=-1$
So,
$y=2e^{-x}+2x-1$
