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

Correct Answer: 1

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