JEE Main 2021 · 1 September · evening · Single correct · +4 / −1
If y = y(x) is the solution curve of the differential equation x²dy + ( y - 1/x )dx = 0 ; x > 0 and y(1) = 1, then y( 1/2 ) is equal to :
- (a) 3/2 - 1/√e
- (b) 3 + 1/√e
- (c) 3 + e
- (d) 3 - e
Answer and solution
Answer (d) 3 - e
Solution
Maths sir · 5 steps
Divide by x² to get a linear equation; its integrating factor e^(-1/x) does the rest.
- Make it linear. Divide by x² dx to get dy/dx + 1/x²y = 1/x³. This has the form dy/dx + Py = Q with P = 1/x².
- Integrating factor. ∫ 1/x² dx = -1/x, so the integrating factor is e^(-1/x). Then y e^(-1/x) = ∫ e^(-1/x)/x³ dx.
- Do the integral. Put t = -1/x, so dt = dx/x² and 1/x = -t. The integral becomes -∫ t eᵗ dt = eᵗ(1 - t) + C, which is e^(-1/x)(1 + 1/x) + C.
y = 1 + 1/x + C e^(1/x)
- Use y(1) = 1. 1 = 1 + 1 + Ce, so C = -1/e. That gives y = 1 + 1/x - e^(1/x - 1).
- So. At x = 1/2: y = 1 + 2 - e²⁻¹ = 3 - e, which is option (d).
The idea to keep. When a dy and dx form has y appearing only to the first power, divide to reach dy/dx + Py = Q. Find the integrating factor first, then fix C with the given point.
Answer from the NTA final answer key. Solution by PrepWiser.