Question

Solution of 2y sinx(dy/dx) = 2sinxcosx - y²cosx, x = π/2, y = 1 is given by

(A) y² = sinx

(B) y = sin²x

(C) y² = cosx + 1 

(D) None of these 

Answer 1

Answer is (A) y² = sinx

On dividing by sin x

2y(dy/dx) + y2cotx = 2cosx

Put y² = v, we get

dv/dx + vcotx = 2cosx

I.F. = e^∫cotxdx = e^{log sinx} = sin x

Therefore, the solution is v sin x = ∫sinx(2cosx)dx + c 

⇒ y² sin x = sin²x + c

When x = π/2, y = 1, then c = 0.

Therefore,

y² = sinx