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

Answer 2

Given the differential equation: \[ \frac{dy}{dx} + y = k \] Multiplying by the integrating factor \( e^x \): \[ y e^x = k \cdot e^x + c \] Using initial condition \( f(0) = e^{-2} \): \[ c = e^{-2} - k \] Thus, the general solution is: \[ y = k + (e^{-2} - k) e^{-x} \] Integrating from \( 0 \) to \( 2 \): \[ k = \int_{0}^{2} \left( k + (e^{-2} - k) e^{-x} \right) dx \] Solving, \[ k = e^{-2} - 1 \] \[ y = (e^{-2} - 1) + e^{-x} \] Evaluating at \( x = 2 \): \[ f(2) = 2e^{-2} - 1, \quad f(0) = e^{-2} \] \[ 2f(0) - f(2) = 1 \]