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 can follow these steps:

### 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 equation to isolate the differentials:
\[ (1+x)y \, dx = -(1-y)x \, dy \]
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 can integrate both sides:
\[ \int \frac{dx}{(1-y)x} = -\int \frac{dy}{(1+x)y} \]
The left side can be integrated as follows:
\[ \int \frac{dx}{(1-y)x} = \frac{1}{1-y} \ln |x| + C_1 \]
And the right side:
\[ -\int \frac{dy}{(1+x)y} = -\frac{1}{1+x} \ln |y| + C_2 \]

### Step 3: Combining the Integrals
Combining the results from both integrals, we have:
\[ \frac{1}{1-y} \ln |x| + \frac{1}{1+x} \ln |y| = C \]
where \(C\) is a constant that combines \(C_1\) and \(C_2\).

### Step 4: Simplifying the Expression
We can rewrite the equation in a more manageable form. By multiplying through by \((1-y)(1+x)\), we can express the relationship between \(x\) and \(y\):
\[ (1+x) \ln |x| + (1-y) \ln |y| = C(1-y)(1+x) \]

### Step 5: Final Form
This gives us a general solution to the differential equation:
\[ \ln |x| + \ln |y| + x = y + C \]
This can be simplified to:
\[ \ln |xy| + x = y + C \]

### Conclusion
Thus, the solution to the differential equation \((1+x)y \, dx + (1-y)x \, dy = 0\) is:
\[ \ln |xy| + x = y + C \]