## University Calculus: Early Transcendentals (3rd Edition)

$y=2e^{-x}+2x-1$
Here, $y=\int (-2e^{-x}+c) dx$ Now, we have $y=2e^{-x}+cx+c'$ Apply the initial conditions $y(0)=1$ We get $c'=-1$ and $y'(0)=0 \implies c=2$ Thus, we have $y=2e^{-x}+2x-1$