JEE Main 2021 · 20 July · morning · Single correct · +4 / −1
Let y = y(x) be the solution of the differential equation eˣ√(1 - y²) dx + ( y/x )dy = 0, y(1) = -1. Then the value of (y(3))² is equal to :
- (a) 1 - 4e³
- (b) 1 - 4e⁶
- (c) 1 + 4e³
- (d) 1 + 4e⁶
Answer and solution
Answer (b) 1 - 4e⁶
Solution
Maths sir · 4 steps
Multiply by x and divide by √(1-y²): the x-terms and y-terms split cleanly apart.
- Separate variables. Multiply the equation by x and divide by √(1-y²). This gives x eˣ dx + y/√(1-y²) dy = 0.
- Integrate each part. By parts, ∫ x eˣ dx = (x-1)eˣ. With t = 1-y², ∫ y/√(1-y²) dy = -√(1-y²).
(x-1)eˣ - √(1-y²) = C
- Use y(1) = -1. At x = 1 the first term is 0. At y = -1, √(1-1) = 0. So C = 0 and √(1-y²) = (x-1)eˣ.
- So. At x = 3, √(1-y²) = 2e³. Squaring gives 1 - y² = 4e⁶, so (y(3))² = 1 - 4e⁶, which is option (b).
The idea to keep. When a term like √(1-y²) multiplies dx, divide through by it to separate the variables. Then use the initial condition to fix C before you plug in the new x.
Answer from the NTA final answer key. Solution by PrepWiser.