JEE Main 2024 · 30 January · morning · Single correct · +4 / −1
Let y=y(x) be the solution of the differential equation sec x d y+{2(1-x) tan x+x(2-x)} d x=0 such that y(0)=2. Then y(2) is equal to:
- (a) 2{sin (2)+1}
- (b) 2
- (c) 1
- (d) 2{1-sin (2)}
Answer and solution
Answer (b) 2
Solution
Maths sir · 4 steps
Do not reach for an integrating factor: the whole dx bracket is the derivative of one product.
- Clear the secant. Multiply through by cos x. Since cos x tan x = sin x, the equation becomes dy/dx = -[2(1-x)sin x + x(2-x)cos x].
- Spot the product rule. Notice that d/dx[x(2-x)] = 2 - 2x = 2(1-x). So the bracket is exactly what you get when you differentiate x(2-x)sin x.
d/dx[x(2-x)sin x] = 2(1-x)sin x + x(2-x)cos x
- Integrate. This gives dy/dx = -d/dx[x(2-x)sin x], so y = -x(2-x)sin x + C. Putting in y(0) = 2 gives C = 2.
- So. y = 2 - x(2-x)sin x. At x = 2 the factor (2-x) is zero, so the whole product vanishes and y(2) = 2. That is option (b).
The idea to keep. Before using an integrating factor, check whether the messy terms are just d/dx(uv). A polynomial times sin x plus its derivative times cos x is a strong hint.
Answer from the NTA final answer key. Solution by PrepWiser.