The solution of the DE `cos x (1 + cosy )dx - sin y (1+ sin x ) dy=0 ` is

Options

A. `1 + sin x cos y = C `
B. ` (1 + sin x ) (1 + cos y ) = C `
C. `sinx cos y + cos x = C `
D. none of these

Explanation

To solve the differential equation \( \cos x (1 + \cos y) \, dx - \sin y (1 + \sin x) \, dy = 0 \), we can start by rearranging the equation into a more manageable form.

### Step 1: Rearranging the Equation
We can rewrite the equation as:
\[ \frac{dy}{dx} = \frac{\cos x (1 + \cos y)}{\sin y (1 + \sin x)} \]

### Step 2: Separating Variables
Next, we can separate the variables \(y\) and \(x\):
\[ \frac{\sin y}{1 + \cos y} \, dy = \frac{\cos x}{1 + \sin x} \, dx \]

### Step 3: Integrating Both Sides
Now, we integrate both sides:
1. For the left side:
\[ \int \frac{\sin y}{1 + \cos y} \, dy \]
We can use the substitution \(u = 1 + \cos y\), then \(du = -\sin y \, dy\). Thus, the integral becomes:
\[ -\int \frac{1}{u} \, du = -\ln |u| + C_1 = -\ln |1 + \cos y| + C_1 \]

2. For the right side:
\[ \int \frac{\cos x}{1 + \sin x} \, dx \]
We can use the substitution \(v = 1 + \sin x\), then \(dv = \cos x \, dx\). Thus, the integral becomes:
\[ \int \frac{1}{v} \, dv = \ln |v| + C_2 = \ln |1 + \sin x| + C_2 \]

### Step 4: Combining the Results
Now we combine the results from both integrals:
\[ -\ln |1 + \cos y| = \ln |1 + \sin x| + C \]
where \(C = C_2 - C_1\).

### Step 5: Exponentiating Both Sides
Exponentiating both sides gives:
\[ |1 + \cos y| = \frac{1}{K |1 + \sin x|} \quad \text{where } K = e^{-C} \]

### Final Solution
Thus, the implicit solution of the differential equation is:
\[ |1 + \cos y| \cdot |1 + \sin x| = K \]