The solution of the differential equation `(1+xy)xdy+(1-xy)ydx=0` is

Options

A. `(1)/(xy)+log((y)/(x))=C`
B. `-xy+log((y)/(x))=C`
C. `-(1)/(xy)+log((y)/(x))=C`
D. `-(1)/(xy)+log((x)/(y))=C`

Explanation

To solve the differential equation \((1 + xy)xdy + (1 - xy)ydx = 0\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ (1 + xy)xdy + (1 - xy)ydx = 0 \]

### Step 2: Expand and rearrange
Expanding the equation gives us:
\[ xdy + x^2y dy + ydx - xy^2 dx = 0 \]
Now, we can group the terms:
\[ xdy + ydx + (x^2y dy - xy^2 dx) = 0 \]

### Step 3: Factor out common terms
We can factor out \(xy\) from the last two terms:
\[ xdy + ydx + xy(xdy - ydx) = 0 \]
This can be rewritten as:
\[ d(xy) + xy(xdy - ydx) = 0 \]

### Step 4: Recognize the total differential
The term \(d(xy)\) represents the total differential of the product \(xy\):
\[ d(xy) + xy(xdy - ydx) = 0 \]

### Step 5: Divide by \(xy^2\)
Now, we divide the entire equation by \(xy^2\):
\[ \frac{d(xy)}{xy^2} + \frac{xdy - ydx}{xy} = 0 \]
This simplifies to:
\[ \frac{d(xy)}{xy^2} + \frac{dy}{y} - \frac{dx}{x} = 0 \]

### Step 6: Integrate both sides
Now we can integrate both sides:
\[ \int \frac{d(xy)}{xy^2} + \int \frac{dy}{y} - \int \frac{dx}{x} = 0 \]
The first integral gives us \(-\frac{1}{xy}\), the second gives us \(\log y\), and the third gives us \(-\log x\):
\[ -\frac{1}{xy} + \log y - \log x = C \]

### Step 7: Simplify the logarithmic terms
Using the property of logarithms \(\log a - \log b = \log\left(\frac{a}{b}\right)\), we can write:
\[ -\frac{1}{xy} + \log\left(\frac{y}{x}\right) = C \]

### Final Result
Thus, the general solution of the differential equation is:
\[ -\frac{1}{xy} + \log\left(\frac{y}{x}\right) = C \]