Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f^{\prime}(x)+f(x)=\int_{0}^{2} f(t) dt$. If $f(0)=e^{-2}$, then $2f(0)-f(2)$ is equal to ____.

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Differential Equations

Let

f : R \to R

be a differentiable function such that

f^{'} (x) + f (x) = \int_{0}^{2} f (t) d t

. If

f (0) = e^{- 2}

, then

2 f (0) - f (2)

is equal to ____.

Answer: 1

Writing

k = \int_{0}^{2} f (t) d t

, the equation becomes

f^{'} (x) + f (x) = k

, a linear equation with solution

f (x) = k + C e^{- x}

. The condition

f (0) = e^{- 2}

gives

C = e^{- 2} - k

. Imposing

\int_{0}^{2} (k + (e^{- 2} - k) e^{- t}) d t = k

leads to

k = e^{- 2} - 1

, hence

C = 1

and

f (x) = e^{- 2} - 1 + e^{- x}

. Then

f (0) = e^{- 2}

and

f (2) = 2 e^{- 2} - 1

, so

2 f (0) - f (2) = 2 e^{- 2} - (2 e^{- 2} - 1) = 1

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Integer answer

Let $y = y (x)$ be the solution curve of the differential equation $(1 + x^{2}) d y + (y - tan^{- 1} x) d x = 0$ , $y (0) = 1$ . Then the value of $y (1)$ is :
JEE Main 2026View solution →
Let $y = y (x)$ be the solution of the differential equation $sec x \frac{dy}{dx} - 2 y = 2 + 3 sin x, x \in (- \frac{π}{2}, \frac{π}{2})$ , $y (0) = - \frac{7}{4}$ . Then $y (\frac{π}{6})$ is equal to:
JEE Main 2026View solution →
Let the solution curve of the differential equation $x d y - y d x = x^{2} + y^{2} d x, x > 0, y (1) = 0$ , be $y = y (x)$ . Then $y (3)$ is equal to
JEE Main 2026View solution →
If $y = y (x)$ satisfies the differential equation
$16 (x + 9 x) (4 + 9 + x) cos y d y = (1 + 2 sin y) d x$ , $x > 0$ and $y (256) = \frac{π}{2}$ , $y (49) = α$ , then $2 sin α$ is equal to :
JEE Main 2026View solution →
Let $y = y (x)$ be the solution of the differential equation
$x^{4} d y + (4 x^{3} y + 2 sin x) d x = 0$ , $x > 0$ , $y (\frac{π}{2}) = 0$ .

Then $π^{4} y (\frac{π}{3})$ is equal to :
JEE Main 2026View solution →
If the solution curve $y = f (x)$ of the differential equation $(x^{2} - 4) y^{'} - 2 x y + 2 x (4 - x^{2})^{2} = 0, x > 2$ , passes through the point (3,15), then the local maximum value of f is
JEE Main 2026View solution →