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

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 \( x = 2 \): \[ f(2) = 2e^{-2} - 1, \quad f(0) = e^{-2} \] \[ 2f(0) - f(2) = 1 \]