JEE Main 2022 · 27 June · evening · Single correct · +4 / −1
If the solution curve of the differential equation
((tan ⁻¹y) - x)dy = (1 + y²)dx passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is
- (a) 2e
- (b) 2/e
- (c) 2
- (d) 1/e
Answer and solution
Answer (b) 2/e
Solution
Maths sir · 4 steps
Flip it to dx/dy: then the equation is linear in x, and tan⁻¹ y becomes the natural new variable.
- Treat x as unknown. Divide by dy and by 1+y². This gives dx/dy + x/(1+y²) = (tan⁻¹ y)/(1+y²), a linear equation in x with y as the variable.
- Integrating factor. The integrating factor is e^(∫ dy/(1+y²)) = e^(tan⁻¹ y). Put t = tan⁻¹ y, so dt = dy/(1+y²) and ∫ t eᵗ dt = (t-1)eᵗ + C by parts.
x e^(tan⁻¹ y) = (tan⁻¹ y - 1)e^(tan⁻¹ y) + C ⇒ x = tan⁻¹ y - 1 + C e^(-tan⁻¹ y)
- Use the point. At (1, 0), tan⁻¹ 0 = 0, so 1 = 0 - 1 + C, which gives C = 2.
- So. When y = tan 1, tan⁻¹ y = 1, so x = 1 - 1 + 2e⁻¹ = 2/e. The answer is (b).
The idea to keep. If dx is multiplied by a function of y alone and x appears only to the first power, write the equation as dx/dy + P(y)x = Q(y) and solve it as linear in x.
Answer from the NTA final answer key. Solution by PrepWiser.