Question

Let $f: R \rightarrow R$ be a differentiable function such that $f^{\prime}(x)+f(x)=\int\limits_0^2 f(t) d t$. If $f(0)=e^{-2}$, then $2 f(0)-f(2)$ is equal to_____

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 =\int\limits_0^2\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)=2 e ^{-2}-1, f (0)= e ^{-2} $
$ 2 f (0)- f (2)=1$