The solution of the differential equation `e^(-x) (y+1) dy +(cos^(2) x - sin 2x) y (dx) = 0` subjected to the condition that y = 1 when x = 0 is

Options

A. `(y+1) + e^(x) cos^(2)x = 2`
B. `y + log y = e^(x) cos^(2) x`
C. `log(y+1)+e^(x) cos^(2)x = 1`
D. `y = log y + e^(x) cos^(2)x = 2`

Explanation

To solve the differential equation \( e^{-x} (y + 1) dy + ( \cos^2 x - \sin 2x ) y dx = 0 \) with the initial condition \( y(0) = 1 \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given equation:
\[ e^{-x} (y + 1) dy + ( \cos^2 x - \sin 2x ) y dx = 0 \]
We can rearrange this to:
\[ e^{-x} (y + 1) dy = - ( \cos^2 x - \sin 2x ) y dx \]

### Step 2: Separate Variables
Next, we can divide both sides by \( y \) and \( e^{-x} \):
\[ \frac{y + 1}{y} dy + ( \cos^2 x - \sin 2x ) dx = 0 \]
This can be rewritten as:
\[ \left( 1 + \frac{1}{y} \right) dy + ( \cos^2 x - \sin 2x ) dx = 0 \]

### Step 3: Integrate Both Sides
Now we can integrate both sides separately:
\[ \int \left( 1 + \frac{1}{y} \right) dy + \int ( \cos^2 x - \sin 2x ) dx = 0 \]

#### Integrating the Left Side
The left side integrates to:
\[ y + \ln |y| + C_1 \]

#### Integrating the Right Side
For the right side, we need to integrate \( \cos^2 x - \sin 2x \):
- Recall that \( \cos^2 x = \frac{1 + \cos 2x}{2} \).
- The integral of \( \sin 2x \) is \( -\frac{1}{2} \cos 2x \).

Thus, we can rewrite:
\[ \int \left( \frac{1 + \cos 2x}{2} - \sin 2x \right) dx = \frac{1}{2} x + \frac{1}{4} \sin 2x + \frac{1}{2} \cos 2x + C_2 \]

### Step 4: Combine the Results
Combining both integrals, we have:
\[ y + \ln |y| = -\left( \frac{1}{2} x + \frac{1}{4} \sin 2x + \frac{1}{2} \cos 2x \right) + C \]

### Step 5: Apply Initial Condition
Now we apply the initial condition \( y(0) = 1 \):
\[ 1 + \ln |1| = -\left( \frac{1}{2}(0) + \frac{1}{4} \sin(0) + \frac{1}{2} \cos(0) \right) + C \]
This simplifies to:
\[ 1 = -\left( 0 + 0 + \frac{1}{2} \right) + C \]
Thus:
\[ 1 = -\frac{1}{2} + C \implies C = \frac{3}{2} \]

### Final Solution
Substituting \( C \) back into the equation gives us:
\[ y + \ln |y| = -\left( \frac{1}{2} x + \frac{1}{4} \sin 2x + \frac{1}{2} \cos 2x \right) + \frac{3}{2} \]

### Summary
The solution to the differential equation is:
\[ y + \ln |y| = -\frac{1}{2} x - \frac{1}{4} \sin 2x - \frac{1}{2} \cos 2x + \frac{3}{2} \]