JEE Main 2022 · 29 July · evening · Single correct · +4 / −1
If the solution curve of the differential equation (d y)/(d x)=(x+y-2)/(x-y) passes through the points (2,1) and (k+1,2), k>0, then
- (a) 2 tan ⁻¹(1/k)=log ₑ(k²+1)
- (b) tan ⁻¹(1/k)=log ₑ(k²+1)
- (c) 2 tan ⁻¹(1/(k+1))=log ₑ(k²+2 k+2)
- (d) 2 tan ⁻¹(1/k)=log ₑ((k²+1)/k²)
Answer and solution
Answer (a) 2 tan ⁻¹(1/k)=log ₑ(k²+1)
Solution
Maths sir · 4 steps
Shift the origin to where the top and bottom both vanish, and this becomes a plain homogeneous equation.
- Shift the origin. Solve x + y - 2 = 0 and x - y = 0 together to get (1, 1). Put x = X + 1, y = Y + 1, and the equation becomes dY/dX = (X + Y)/(X - Y), which is homogeneous.
- Substitute Y = vX. Then v + Xdv/dX = (1 + v)/(1 - v), so Xdv/dX = (1 + v²)/(1 - v). Separate the variables and integrate: ∫ (1 - v)/(1 + v²) dv = ∫ dX/X.
tan⁻¹ v - 1/2 ln(1 + v²) = ln|X| + C ⇒ 2 tan⁻¹ Y/X = ln(X² + Y²) + C'
- Use the first point. (2, 1) gives X = 1, Y = 0. Then 2 tan⁻¹ 0 = ln 1 + C', so C' = 0.
- So. (k + 1, 2) gives X = k, Y = 1. This leads to 2 tan⁻¹(1/k) = ln(k² + 1), which is option (a).
The idea to keep. When dy/dx is a ratio of linear expressions with constant terms, first find where both lines meet. Move the origin there and the constants disappear.
Answer from the NTA final answer key. Solution by PrepWiser.