JEE Main 2023 (Online) 1st February Morning ShiftQuestion 1
Let $f:\mathbb R\to\mathbb R$ be a differentiable function such that $f'(x)+f(x)=\int_0^2f(t)dt$. If $f(0)=e^{-2}$, then $2f(0)-f(2)$ is equal to ______.
Differential Equation
Answer$1$
Solution
Concept. The integral is a constant $k$; solve $f'+f=k$, then find $k$ from its definition. Formulas used. (1) $f'+f=k\Rightarrow f=k+(f(0)-k)\,e^{-x}$Step 1 - Solve. By (1), $f=k+(e^{-2}-k)e^{-x}$. Step 2 - Find $k$. $k=\int_0^2f\,dt=2k+(e^{-2}-k)(1-e^{-2})$. This simplifies to $$0=e^{-2}-e^{-4}+ke^{-2}\ \Longrightarrow\ k=e^{-2}-1.$$ Step 3 - $f(2)$. $e^{-2}-k=1$, so $f(2)=k+e^{-2}=2e^{-2}-1$. Step 4 - Value. $2f(0)-f(2)=2e^{-2}-2e^{-2}+1=1$. Answer: 1