The general solution of the differential equation `(dy)/(dx)=2y tan x+tan^(2)x, AA x in (0, (pi)/(2))` is `yf(x)=(x)/(2)-(sin(2x))/(4)+C`, (where, C is an arbitrary constant). If `f((pi)/(4))=(1)/(2)`, then the value of `f((pi)/(3))` is equal to

Options

A. `(1)/(2)`
B. `(1)/(4)`
C. `2`
D. `4`

Answer

To solve the problem step by step, we start with the given differential equation and the general solution provided.

### Step 1: Understanding the Differential Equation
The differential equation given is:
\[ \frac{dy}{dx} = 2y \tan x + \tan^2 x \]
We need to find the general solution of this equation.

### Step 2: Rearranging the Equation
We can rearrange the equation as follows:
\[ \frac{dy}{dx} - 2y \tan x = \tan^2 x \]
This is in the standard form of a first-order linear differential equation:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
where \( P(x) = -2 \tan x \) and \( Q(x) = \tan^2 x \).

### Step 3: Finding the Integrating Factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int -2 \tan x \, dx} \]
Calculating the integral:
\[ \int -2 \tan x \, dx = -2 \ln |\sec x| = \ln |\sec^2 x|^{-1} = \ln |\cos^2 x| \]
Thus, the integrating factor is:
\[ \mu(x) = e^{\ln |\cos^2 x|} = \cos^2 x \]

### Step 4: Multiplying the Equation by the Integrating Factor
Now, we multiply the entire differential equation by \( \cos^2 x \):
\[ \cos^2 x \frac{dy}{dx} - 2y \cos^2 x \tan x = \cos^2 x \tan^2 x \]
This simplifies to:
\[ \frac{d}{dx}(y \cos^2 x) = \cos^2 x \tan^2 x \]

### Step 5: Integrating Both Sides
Integrating both sides:
\[ \int \frac{d}{dx}(y \cos^2 x) \, dx = \int \cos^2 x \tan^2 x \, dx \]
The left side becomes:
\[ y \cos^2 x = \int \cos^2 x \tan^2 x \, dx + C \]

### Step 6: Solving the Integral
Using the identity \( \tan^2 x = \sec^2 x - 1 \):
\[ \int \cos^2 x \tan^2 x \, dx = \int \cos^2 x (\sec^2 x - 1) \, dx = \int (1 - \cos^2 x) \, dx \]
This results in:
\[ \int (1 - \cos^2 x) \, dx = x - \frac{1}{2} \sin(2x) + C \]

### Step 7: General Solution
Thus, we have:
\[ y \cos^2 x = x - \frac{1}{2} \sin(2x) + C \]
Rearranging gives:
\[ y = \frac{x}{\cos^2 x} - \frac{\sin(2x)}{2 \cos^2 x} + C \]

### Step 8: Finding \( f(x) \)
From the given general solution:
\[ y f(x) = \frac{x}{2} - \frac{\sin(2x)}{4} + C \]
We can identify:
\[ f(x) = \cos^2 x \]

### Step 9: Using the Condition
We are given \( f\left(\frac{\pi}{4}\right) = \frac{1}{2} \):
\[ f\left(\frac{\pi}{4}\right) = \cos^2\left(\frac{\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
This confirms our function \( f(x) \).

### Step 10: Finding \( f\left(\frac{\pi}{3}\right) \)
Now we need to find:
\[ f\left(\frac{\pi}{3}\right) = \cos^2\left(\frac{\pi}{3}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

### Final Answer
Thus, the value of \( f\left(\frac{\pi}{3}\right) \) is:
\[ \frac{1}{4} \]