JEE Main 2024 · 9 April · morning · Single correct · +4 / −1
The solution of the differential equation (x²+y²) d x-5 x y d y=0, y(1)=0, is :
- (a) |x²-4 y²|⁵=x²
- (b) |x²-2 y²|⁶=x
- (c) |x²-2 y²|⁵=x²
- (d) |x²-4 y²|⁶=x
Answer and solution
Answer (a) |x²-4 y²|⁵=x²
Solution
Maths sir · 4 steps
Every term has degree 2, so put y = vx and the equation separates neatly in v and x.
- Substitute y = vx. Write dy/dx = (x²+y²)/5xy. With y = vx and dy/dx = v + xdv/dx, the right side becomes (1+v²)/5v.
- Separate. xdv/dx = (1+v²)/5v - v = (1-4v²)/5v. So 5v/(1-4v²) dv = dx/x.
-5/8 ln|1-4v²| = ln|x| + C
- Clear the logs. Multiply by -8: |1-4v²|⁵ = c/x⁸. Put 1-4v² = (x²-4y²)/x² to get |x²-4y²|⁵ = c x¹⁰/x⁸ = c x².
- So. At x = 1, y = 0 we get 1 = c. The solution is |x²-4y²|⁵ = x², option (a).
The idea to keep. When every term has the same degree, put y = vx. Then always simplify v + xdv/dx by moving v across first. That subtraction decides the coefficients in the final answer.
Answer from the NTA final answer key. Solution by PrepWiser.