JEE Main 2021 · 26 August · evening · Single correct · +4 / −1
Let y(x) be the solution of the differential equation
2x² dy + (eʸ - 2x)dx = 0, x > 0. If y(e) = 1, then y(1) is equal to :
- (a) 0
- (b) 2
- (c) logₑ 2
- (d) logₑ (2e)
Answer and solution
Answer (c) logₑ 2
Solution
Maths sir · 4 steps
The eʸ term spoils linearity, so put t = e⁻ʸ and the equation becomes linear in t.
- Rearrange. Divide by 2x² dx to get dy/dx = 1/x - eʸ/2x². Multiply through by e⁻ʸ: e⁻ʸdy/dx = e⁻ʸ/x - 1/2x².
- Substitute t. Let t = e⁻ʸ, so dt/dx = -e⁻ʸdy/dx. The equation becomes dt/dx + t/x = 1/2x², which is linear in t with integrating factor e^(∫ 1/xdx) = x.
d/dx(x e⁻ʸ) = 1/2x ⇒ x e⁻ʸ = 1/2 ln x + C
- Find C. Put x = e, y = 1: e · e⁻¹ = 1/2 + C, so 1 = 1/2 + C and C = 1/2.
- So. At x = 1: e⁻ʸ = 0 + 1/2, so eʸ = 2 and y(1) = logₑ 2, option (c).
The idea to keep. When a function of y like eʸ or ln y blocks linearity, divide so that its derivative appears, then substitute. The equation turns linear in the new variable.
Answer from the NTA final answer key. Solution by PrepWiser.