Find the particular solution of `cosy dx+(1+ 2e^(-x)) sin y dy =0` when `x=0, y=pi/4`

Options

A. If both assertion and reason are CORRECT and the reason is CORRECT explanation of the assertion.
B. If both assertion and reason are CORRECT and the reason in INCORRECT explanation of the assertion.
C. If assertion is CORRECT and the reason in INCORRECT
D. If assertion in INCORRECT and the reason is CORRECT.

Answer

To find the particular solution of the differential equation

\[ \cos y \, dx + (1 + 2e^{-x}) \sin y \, dy = 0 \]

given the initial condition \(x = 0\) and \(y = \frac{\pi}{4}\), we can follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation:

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

Now, we can separate the variables:

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

This simplifies to:

\[ \frac{dx}{1 + 2e^{-x}} = -\tan y \, dy \]

### Step 2: Integrating Both Sides
Next, we integrate both sides. The left side requires a substitution, while the right side integrates directly:

\[ \int \frac{dx}{1 + 2e^{-x}} = -\int \tan y \, dy \]

For the left side, we can use the substitution \(u = 1 + 2e^{-x}\), which gives us:

\[ du = -2e^{-x} \, dx \quad \Rightarrow \quad dx = -\frac{du}{2e^{-x}} = -\frac{du}{2(u - 1)} \]

The integral becomes:

\[ -\frac{1}{2} \int \frac{du}{u} = -\frac{1}{2} \ln |u| + C_1 = -\frac{1}{2} \ln |1 + 2e^{-x}| + C_1 \]

For the right side, we have:

\[ -\int \tan y \, dy = -\ln |\sec y| + C_2 \]

### Step 3: Equating the Integrals
Now we equate the two integrals:

\[ -\frac{1}{2} \ln |1 + 2e^{-x}| = -\ln |\sec y| + C \]

Multiplying through by -2 gives:

\[ \ln |1 + 2e^{-x}| = 2\ln |\sec y| - 2C \]

### Step 4: Exponentiating Both Sides
Exponentiating both sides results in:

\[ 1 + 2e^{-x} = K \sec^2 y \quad \text{where } K = e^{-2C} \]

### Step 5: Finding the Particular Solution
Now we apply the initial conditions \(x = 0\) and \(y = \frac{\pi}{4}\):

\[ 1 + 2e^{0} = K \sec^2\left(\frac{\pi}{4}\right) \]

This simplifies to:

\[ 1 + 2 = K \cdot 2 \quad \Rightarrow \quad 3 = 2K \quad \Rightarrow \quad K = \frac{3}{2} \]

### Step 6: Final Equation
Substituting \(K\) back into our equation gives:

\[ 1 + 2e^{-x} = \frac{3}{2} \sec^2 y \]

This can be rearranged to find the particular solution:

\[ 2e^{-x} = \frac{3}{2} \sec^2 y - 1 \]

\[ e^{-x} = \frac{3 \sec^2 y - 2}{4} \]

### Final Answer
Thus, the particular solution is:

\[ e^{-x} = \frac{3 \sec^2 y - 2}{4} \]