Application Of Derivatives Question 12
Question 12 - 01 February - Shift 1
Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f^{\prime}(x)+f(x)=\int_0^{2} f(t) d t$. If $f(0)=e^{-2}$, then $2 f(0)-f(2)$ is equal to
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Answer: 1
Solution:
Formula: Linear Differential Equations Of First Order (7.1), Properties of definite integral
$ \begin{aligned} & \frac{dy}{dx}+y=k \\ & \quad y \cdot e^{x}=k \cdot e^{x}+c \\ & f(0)=e^{-2} \\ & \Rightarrow \quad c=e^{-2}-k \\ & \therefore \quad y=k+(e^{-2}-k) e^{-x} \\ & \\ & \quad \text{ now } k=\int_0^{2}(k+(e^{-2}-k) e^{-x}) dx \\ & \Rightarrow \quad k=e^{-2}-1 \\ & \therefore \quad y=(e^{-2}-1)+e^{-x} \\ & \\ & f(2)=2 e^{-2}-1, f(0)=e^{-2} \\ & 2 f(0)-f(2)=1 \end{aligned} $
