If `dy/dx-(sin2x)/(1+cos^2x)y=(sinx)/(1+cos^2x)` and `y(0)=0` then `y(pi/2)` is

Options

A. 2
B. 1
C. 5
D. 3

Answer

To solve the differential equation

\[ \frac{dy}{dx} - \frac{\sin 2x}{1 + \cos^2 x} y = \frac{\sin x}{1 + \cos^2 x} \]

with the initial condition \( y(0) = 0 \), we can follow these steps:

### Step 1: Identify the components of the linear differential equation
The equation is in the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \), where:
- \( P(x) = -\frac{\sin 2x}{1 + \cos^2 x} \)
- \( Q(x) = \frac{\sin x}{1 + \cos^2 x} \)

### Step 2: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int -\frac{\sin 2x}{1 + \cos^2 x} \, dx} \]

To compute this integral, we can use the substitution \( t = 1 + \cos^2 x \). Then, we have:

\[ dt = -2\cos x \sin x \, dx \quad \Rightarrow \quad -\sin 2x \, dx = dt \]

Thus,

\[ \int -\frac{\sin 2x}{1 + \cos^2 x} \, dx = \int \frac{dt}{t} = \ln |t| + C = \ln |1 + \cos^2 x| + C \]

So, the integrating factor is:

\[ \mu(x) = e^{\ln(1 + \cos^2 x)} = 1 + \cos^2 x \]

### Step 3: Multiply the differential equation by the integrating factor
Multiply the entire differential equation by \( 1 + \cos^2 x \):

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

This simplifies to:

\[ (1 + \cos^2 x) \frac{dy}{dx} - \sin 2x \cdot y = \sin x \]

### Step 4: Rewrite the left-hand side
The left-hand side can be written as the derivative of a product:

\[ \frac{d}{dx} \left( y(1 + \cos^2 x) \right) = \sin x \]

### Step 5: Integrate both sides
Integrate both sides with respect to \( x \):

\[ y(1 + \cos^2 x) = \int \sin x \, dx = -\cos x + C \]

### Step 6: Solve for \( y \)
Rearranging gives:

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

Thus,

\[ y = \frac{-\cos x + C}{1 + \cos^2 x} \]

### Step 7: Apply the initial condition
Using the initial condition \( y(0) = 0 \):

\[ 0 = \frac{-\cos(0) + C}{1 + \cos^2(0)} = \frac{-1 + C}{2} \]

This implies:

\[ C = 1 \]

### Step 8: Substitute \( C \) back into the equation for \( y \)
Now we have:

\[ y = \frac{-\cos x + 1}{1 + \cos^2 x} \]

### Step 9: Find \( y(\frac{\pi}{2}) \)
Now, we need to find \( y(\frac{\pi}{2}) \):

\[ y\left(\frac{\pi}{2}\right) = \frac{-\cos\left(\frac{\pi}{2}\right) + 1}{1 + \cos^2\left(\frac{\pi}{2}\right)} = \frac{-0 + 1}{1 + 0} = 1 \]

Thus, the final answer is:

\[ \boxed{1} \]