A function y=f (x) satisfies the differential equation `f(x) sin 2x-cos x+(1 + sin^2x)f'(x)= 0` with initial condition `y(0)=0`. The value of `f(pi/6)` is equal to (A) `1/5` (B) `3/5` (C) `4/5` (D) `2/5`

Options

A. `1/5`
B. `3/5`
C. `4/5`
D. `2/5`

Answer

To solve the given differential equation \( f(x) \sin 2x - \cos x + (1 + \sin^2 x) f'(x) = 0 \) with the initial condition \( f(0) = 0 \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the equation:
\[ f(x) \sin 2x - \cos x + (1 + \sin^2 x) f'(x) = 0 \]
Rearranging gives:
\[ (1 + \sin^2 x) f'(x) = \cos x - f(x) \sin 2x \]
Now, we can express this in the standard form of a linear differential equation:
\[ f'(x) + \frac{\sin 2x}{1 + \sin^2 x} f(x) = \frac{\cos x}{1 + \sin^2 x} \]

### Step 2: Identify \( p(x) \) and \( q(x) \)
From the standard form \( f' + p(x) f = q(x) \), we identify:
\[ p(x) = \frac{\sin 2x}{1 + \sin^2 x}, \quad q(x) = \frac{\cos x}{1 + \sin^2 x} \]

### Step 3: Find the Integrating Factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int p(x) \, dx} = e^{\int \frac{\sin 2x}{1 + \sin^2 x} \, dx} \]
Let \( t = 1 + \sin^2 x \), then \( dt = 2 \sin x \cos x \, dx = \sin 2x \, dx \). Thus:
\[ \int \frac{\sin 2x}{1 + \sin^2 x} \, dx = \int \frac{dt}{t} = \ln |t| + C = \ln |1 + \sin^2 x| + C \]
So, the integrating factor is:
\[ \mu(x) = 1 + \sin^2 x \]

### Step 4: Multiply through by the Integrating Factor
Multiply the entire differential equation by the integrating factor:
\[ (1 + \sin^2 x) f' + \sin 2x f = \cos x \]

### Step 5: Integrate Both Sides
The left-hand side can be expressed as the derivative of a product:
\[ \frac{d}{dx} \left( (1 + \sin^2 x) f \right) = \cos x \]
Integrating both sides gives:
\[ (1 + \sin^2 x) f = \int \cos x \, dx = \sin x + C \]

### Step 6: Solve for \( f(x) \)
Thus, we have:
\[ f(x) = \frac{\sin x + C}{1 + \sin^2 x} \]

### Step 7: Apply the Initial Condition
Using the initial condition \( f(0) = 0 \):
\[ f(0) = \frac{\sin 0 + C}{1 + \sin^2 0} = \frac{0 + C}{1 + 0} = C \]
Since \( f(0) = 0 \), we find \( C = 0 \). Therefore:
\[ f(x) = \frac{\sin x}{1 + \sin^2 x} \]

### Step 8: Find \( f\left(\frac{\pi}{6}\right) \)
Now we compute:
\[ f\left(\frac{\pi}{6}\right) = \frac{\sin\left(\frac{\pi}{6}\right)}{1 + \sin^2\left(\frac{\pi}{6}\right)} = \frac{\frac{1}{2}}{1 + \left(\frac{1}{2}\right)^2} = \frac{\frac{1}{2}}{1 + \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{5}{4}} = \frac{1}{2} \cdot \frac{4}{5} = \frac{2}{5} \]

### Final Answer
Thus, the value of \( f\left(\frac{\pi}{6}\right) \) is:
\[ \boxed{\frac{2}{5}} \]