A function y=f(x) satisfies f(x) sin 2 x+sin x-(1+cos ² x) f′(x)=0 with condition f(0)=0. Then, f(π/2) is equal to

Solution

Maths sir · 4 steps

Spot that -sin 2x is the derivative of 1+cos² x, and the whole left side becomes one derivative.

1. Standard linear form. Rearrange to (1+cos² x) y' - y sin 2x = sin x. Divide by 1+cos² x to get y' + P y = Q with P = -(sin 2x)/(1+cos² x) and Q = (sin x)/(1+cos² x).
2. Integrating factor. Since d/dx(1+cos² x) = -2 cos x sin x = -sin 2x, we get ∫ P dx = ln(1+cos² x). So the integrating factor is e^(∫ P dx) = 1+cos² x.

   y (1+cos² x) = ∫ sin x dx = -cos x + C

3. Use f(0) = 0. At x = 0: 0 × 2 = -1 + C, so C = 1. This gives f(x) = (1-cos x)/(1+cos² x).
4. So. f(π/2) = (1-0)/(1+0) = 1, which is option (b).

The idea to keep. Before hunting for an integrating factor, check whether the coefficient of y is the derivative of the coefficient of y'. If it is, the left side is already d/dx of a product.

Answer from the NTA final answer key. Solution by PrepWiser.