Solution of the differential equation `(1+x)y dx+(1-y)x dy=0` is

Options

A. `log(x*y)+x-y=c`.
B. `log(x/y)+x-y=c`.
C. `log(x*y)+x=c`.
D. none of these

Explanation

To solve the differential equation \((1+x)y \, dx + (1-y)x \, dy = 0\), we will use the method of separation of variables. Here’s a step-by-step solution:

### Step 1: Rearranging the Equation
We start with the given differential equation:
\[ (1+x)y \, dx + (1-y)x \, dy = 0 \]
We can rearrange this to separate the variables \(x\) and \(y\):
\[ (1+x)y \, dx = -(1-y)x \, dy \]
Dividing both sides by \((1-y)x(1+x)y\) gives:
\[ \frac{dx}{(1-y)x} = -\frac{dy}{(1+x)y} \]

### Step 2: Integrating Both Sides
Now we will integrate both sides:
\[ \int \frac{dx}{(1-y)x} = -\int \frac{dy}{(1+x)y} \]
The left side integrates to:
\[ \int \frac{dx}{(1-y)x} = \frac{1}{1-y} \log |x| + C_1 \]
The right side integrates to:
\[ -\int \frac{dy}{(1+x)y} = -\frac{1}{1+x} \log |y| + C_2 \]

### Step 3: Combining the Integrals
Now we can combine the results of the integrals:
\[ \frac{1}{1-y} \log |x| + \frac{1}{1+x} \log |y| = C \]
where \(C\) is a constant that combines \(C_1\) and \(C_2\).

### Step 4: Simplifying the Equation
We can rewrite the equation:
\[ \log |x| + \log |y| = C(1-y)(1+x) \]
Using the property of logarithms, we can combine the logs:
\[ \log |xy| = C(1-y)(1+x) \]

### Step 5: Final Form
Exponentiating both sides gives us:
\[ xy = e^{C(1-y)(1+x)} \]
This is the implicit solution of the differential equation.