Question

Let $f:R\to R$ be a differentiable function such that $f^{\prime }(x)+f(x)=\underset{0}{\overset{2}{\int }}f(t)dt$. If $f(0)=e^{-2}$, then $2f(0)-f(2)$ is equal to_____

Answer 1

Answer: 1

$\frac{dy}{dx}+y=k$
$y\cdot e^x=k\cdot e^x+c$
$f(0)=e^{-2}$
$\Rightarrow c=e^{-2}-k$
$\therefore y=k+\left(e^{-2}-k\right)e^{-x}$
$\text{ now }k=\underset{0}{\overset{2}{\int }}\left(k+\left(e^{-2}-k\right)e^{-x}\right)dx$
$\Rightarrow k=e^{-2}-1$
$\therefore y=\left(e^{-2}-1\right)+e^{-x}$
$f(2)=2e^{-2}-1,f(0)=e^{-2}$
$2f(0)-f(2)=1$