Exams › JEE Main › Maths
Let y = y(x) be the solution of the differential equation dy/dx + 2y = f(x), where f(x) = { 1, x ∈ [0,1]; 0, otherwise } If y(0) = 0, then (3/2) is
- (e² - 1)/(2e³)
- (e² - 1)/e³
- 1/(2e)
- (e² + 1)/(2e⁴)
Correct answer: (e² - 1)/(2e³)
Solution
On [0,1], dy/dx+2y=1 with y(0)=0 gives y=(1/2)(1-e^(-2x)), so y(1)=(1/2)(1-e^-2). For x>1, dy/dx+2y=0 so y(3/2)=y(1)e^-1 = (1/2)(e^-1-e^-3) = (e^2-1)/(2e^3).
Related JEE Main Maths questions
- If √(1-x²ⁿ)+√(1-y²ⁿ)=a(xⁿ-yⁿ), then the value of (√(1-x²ⁿ) dy)/(√(1-y²ⁿ) dx) is
- For a curve that passes through the point (4, 0), the slope is governed by
dy/dx = y/x + 5x/((x + 2)(x − 3)).
If the point (5, a) lies on this curve, what is the value of a?
- Which differential equation represents the family of all conics whose axes are aligned with the coordinate axes?
- Find the equation of the curve that satisfies (xy - x²) (dy)/(dx) = y² and passes through the point (-1, 1).
- For the differential equation y = y/x + x/y, if its general solution is written as y = x / log|Cx|, then the function φ(x/y) is
- For the differential equation dy/dx = [y f'(x) − y²]/f(x), where f(x) is a specified function, the solution is
⚔️ Practice JEE Main Maths free + battle 1v1 →