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

Answer: (A)

1

Answer 2

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