The solution of the differential equation `(dy)/(dx) = (xy+y)/(xy+x)` is

Options

A. `x+y = log((Cy)/x)`
B. `x+y = log(Cxy)`
C. `x-y = log ((Cx)/y)`
D. ` y - x = log ((Cx)/y)`

Explanation

To solve the differential equation \(\frac{dy}{dx} = \frac{xy + y}{xy + x}\), we can follow these steps:

### Step 1: Simplify the Equation
We start with the given equation:
\[ \frac{dy}{dx} = \frac{xy + y}{xy + x} \]
We can factor out \(y\) from the numerator and \(x\) from the denominator:
\[ \frac{dy}{dx} = \frac{y(x + 1)}{x(y + 1)} \]

### Step 2: Separate Variables
Now we can separate the variables:
\[ \frac{dy}{y + 1} = \frac{x + 1}{x} dx \]

### Step 3: Integrate Both Sides
Next, we integrate both sides:
\[ \int \frac{dy}{y + 1} = \int \left(1 + \frac{1}{x}\right) dx \]
The left side integrates to:
\[ \ln |y + 1| \]
The right side integrates to:
\[ x + \ln |x| + C \]
So we have:
\[ \ln |y + 1| = x + \ln |x| + C \]

### Step 4: Exponentiate Both Sides
To eliminate the logarithm, we exponentiate both sides:
\[ |y + 1| = e^{x + \ln |x| + C} \]
This simplifies to:
\[ |y + 1| = e^C \cdot |x| \cdot e^x \]
Let \(K = e^C\), then:
\[ |y + 1| = K |x| e^x \]

### Step 5: Solve for \(y\)
Now we can solve for \(y\):
\[ y + 1 = K x e^x \quad \text{or} \quad y + 1 = -K x e^x \]
Thus, we can express \(y\) as:
\[ y = K x e^x - 1 \quad \text{or} \quad y = -K x e^x - 1 \]

### Final Solution
The general solution of the differential equation is:
\[ y = C x e^x - 1 \]
where \(C\) is a constant.

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