Let y = y (x) be the solution of the differential equation
` cos x (dy)/(dx) + 2y sin x = sin 2x , x in (0, pi/2) ` .
If `y(pi//3) = 0, " then " y(pi//4)` is equal to :

Options

A. `2 - sqrt2`
B. `2 + sqrt2`
C. `sqrt2 - 2`
D. `(1)/(sqrt2) - 1`

Answer

To solve the given differential equation \( \cos x \frac{dy}{dx} + 2y \sin x = \sin 2x \) with the initial condition \( y\left(\frac{\pi}{3}\right) = 0 \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the equation:
\[ \cos x \frac{dy}{dx} + 2y \sin x = \sin 2x \]
Dividing through by \( \cos x \) gives:
\[ \frac{dy}{dx} + 2y \tan x = \frac{\sin 2x}{\cos x} \]

### Step 2: Identify \( p(x) \) and \( q(x) \)
From the standard form \( \frac{dy}{dx} + p(x)y = q(x) \), we identify:
\[ p(x) = 2 \tan x, \quad q(x) = \frac{\sin 2x}{\cos x} \]

### Step 3: Find the Integrating Factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int p(x) \, dx} = e^{\int 2 \tan x \, dx} \]
The integral of \( \tan x \) is \( -\ln(\cos x) \), thus:
\[ \mu(x) = e^{2 \ln(\sec x)} = \sec^2 x \]

### Step 4: Multiply the Equation by the Integrating Factor
Multiplying the entire differential equation by \( \sec^2 x \):
\[ \sec^2 x \frac{dy}{dx} + 2y \sec^2 x \tan x = 2 \sec^2 x \sin x \]

### Step 5: Recognize the Left Side as a Derivative
The left side can be expressed as the derivative of a product:
\[ \frac{d}{dx}(y \sec^2 x) = 2 \sec^2 x \sin x \]

### Step 6: Integrate Both Sides
Integrating both sides gives:
\[ y \sec^2 x = \int 2 \sec^2 x \sin x \, dx \]
Using integration by parts or recognizing the integral:
\[ \int 2 \sec^2 x \sin x \, dx = 2 \sin x \tan x - 2 \int \tan x \cos x \, dx \]
This simplifies to:
\[ y \sec^2 x = 2 \sin x \tan x + C \]

### Step 7: Solve for \( y \)
Rearranging gives:
\[ y = 2 \sin x \tan x \cos^2 x + C \cos^2 x \]

### Step 8: Apply the Initial Condition
Using the initial condition \( y\left(\frac{\pi}{3}\right) = 0 \):
\[ 0 = 2 \sin\left(\frac{\pi}{3}\right) \tan\left(\frac{\pi}{3}\right) \cos^2\left(\frac{\pi}{3}\right) + C \cos^2\left(\frac{\pi}{3}\right) \]
Calculating the values:
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \quad \tan\left(\frac{\pi}{3}\right) = \sqrt{3}, \quad \cos^2\left(\frac{\pi}{3}\right) = \frac{1}{4} \]
Substituting gives:
\[ 0 = 2 \cdot \frac{\sqrt{3}}{2} \cdot \sqrt{3} \cdot \frac{1}{4} + C \cdot \frac{1}{4} \]
This simplifies to:
\[ 0 = \frac{3}{4} + \frac{C}{4} \implies C = -3 \]

### Step 9: Substitute Back to Find \( y \)
Thus, the equation becomes:
\[ y = 2 \sin x \tan x \cos^2 x - 3 \cos^2 x \]

### Step 10: Find \( y\left(\frac{\pi}{4}\right) \)
Calculating \( y\left(\frac{\pi}{4}\right) \):
\[ y\left(\frac{\pi}{4}\right) = 2 \sin\left(\frac{\pi}{4}\right) \tan\left(\frac{\pi}{4}\right) \cos^2\left(\frac{\pi}{4}\right) - 3 \cos^2\left(\frac{\pi}{4}\right) \]
Using \( \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \tan\left(\frac{\pi}{4}\right) = 1, \cos^2\left(\frac{\pi}{4}\right) = \frac{1}{2} \):
\[ y\left(\frac{\pi}{4}\right) = 2 \cdot \frac{1}{\sqrt{2}} \cdot 1 \cdot \frac{1}{2} - 3 \cdot \frac{1}{2} \]
This simplifies to:
\[ y\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} - \frac{3}{2} \]

### Final Answer
Thus, the value of \( y\left(\frac{\pi}{4}\right) \) is:
\[ \frac{1}{\sqrt{2}} - 2 \]