JEE Main 2020 · 5 September · morning · Single correct · +4 / −1
If y = y(x) is the solution of the differential
equation (5 + eˣ)/(2 + y).dy/dx + eˣ = 0 satisfying y(0) = 1, then a value of y(logₑ13) is :
- (a) -1
- (b) 1
- (c) 0
- (d) 2
Answer and solution
Answer (a) -1
Solution
Maths sir · 5 steps
Get all the y on one side and all the x on the other, and both sides become log integrals.
- Separate variables. Move eˣ to the right and rearrange so y and x sit on different sides: dy/(2+y) = -eˣ/(5+eˣ) dx.
- Integrate both sides. The top of each fraction is the derivative of its bottom, so both sides integrate to a log. This gives ln(2+y) = -ln(5+eˣ) + C, which tidies into one product.
(2+y)(5+eˣ) = k
- Initial condition. At x = 0, e⁰ = 1 and y = 1, so (2+1)(5+1) = k and k = 18.
- Plug in the point. At x = logₑ 13, e^(logₑ 13) = 13, so (2+y)(5+13) = 18. That gives 2+y = 1.
- So. y(logₑ 13) = -1, which is option (a).
The idea to keep. When a fraction has its own derivative on top, its integral is a log. Two log terms then combine into a product equal to a constant, and the initial condition fixes that constant.
Answer from the NTA final answer key. Solution by PrepWiser.