Question

A function \(y=f(x)\) satisfies\(f (x)sin2x+sinx-(1+cos^2x) f'(x)=0\) with condition\(f(0)=0\).Then\(f(\frac{\pi}{2})\) equals to

• A. 1
• B. 0
• C. -1
• D. 2

Answer 1

The Correct Option is A

Step 1: Rewrite the Differential Equation

\(\frac{dy}{dx} - \left( \frac{\sin 2x}{1 + \cos^2 x} \right) y = \sin x\)

Step 2: Find the Integrating Factor (I.F.)

The integrating factor is:

I.F. = \(1 + \cos^2 x\)

Step 3: Solve the Differential Equation

Multiply through by the integrating factor:

\(y \cdot (1 + \cos^2 x) = \int (\sin x) dx\)

Integrate:

\(y \cdot (1 + \cos^2 x) = -\cos x + C\)

Step 4: Apply Initial Condition \(f(0) = 0\)

At \(x = 0\):

\(-\cos 0 + C = 0 \Rightarrow C = 1\)

Step 5: Evaluate \(y \left( \frac{\pi}{2} \right)\)

\(y \left( \frac{\pi}{2} \right) = 1\)

So, the correct answer is: 1