## Calculus 10th Edition

$y=Ce^x-10$
$y'-y=10$ Here, $P(x)=-1$ and $Q(x)=10$ Therefore, $u(x)=e^{\int-1dx}=e^{-x}$ So, $y=\frac{1}{e^{-x}}\int 10 e^{-x}dx$ Or, $y=e^x(-10e^{-x}+C)=-10+Ce^x$