What is the general solution of the differential equation `x dy - y dx = y^(2)dx` ?
Where C is an arbitrary constant

Options

A. A. x = Cy
B. B. `y = C-x`
C. C. x + xy - Cy = 0
D. D. None of these

Answer

To solve the differential equation \( x \, dy - y \, dx = y^2 \, dx \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ x \, dy - y \, dx = y^2 \, dx \]

### Step 2: Divide by \( dx \)
Next, we divide both sides by \( dx \):
\[ x \frac{dy}{dx} - y = y^2 \]

### Step 3: Rearrange the equation
Now, we rearrange the equation to isolate the terms involving \( y \):
\[ x \frac{dy}{dx} = y + y^2 \]

### Step 4: Move \( y^2 \) to the left side
We can express it as:
\[ x \frac{dy}{dx} - y - y^2 = 0 \]

### Step 5: Factor out \( y \)
We can factor out \( y \) from the left side:
\[ x \frac{dy}{dx} - y(1 + y) = 0 \]

### Step 6: Isolate \( \frac{dy}{dx} \)
Rearranging gives:
\[ x \frac{dy}{dx} = y(1 + y) \]
Dividing both sides by \( x \):
\[ \frac{dy}{dx} = \frac{y(1 + y)}{x} \]

### Step 7: Separate variables
Now we can separate the variables:
\[ \frac{dy}{y(1 + y)} = \frac{dx}{x} \]

### Step 8: Integrate both sides
Next, we integrate both sides:
\[ \int \frac{dy}{y(1 + y)} = \int \frac{dx}{x} \]

### Step 9: Solve the left integral
Using partial fraction decomposition, we can rewrite the left side:
\[ \frac{1}{y(1+y)} = \frac{A}{y} + \frac{B}{1+y} \]
Solving gives \( A = 1 \) and \( B = -1 \):
\[ \int \left( \frac{1}{y} - \frac{1}{1+y} \right) dy = \int \frac{dx}{x} \]
Integrating, we get:
\[ \ln |y| - \ln |1+y| = \ln |x| + C \]

### Step 10: Combine logarithms
Combining the logarithms:
\[ \ln \left| \frac{y}{1+y} \right| = \ln |x| + C \]

### Step 11: Exponentiate both sides
Exponentiating both sides gives:
\[ \frac{y}{1+y} = Kx \quad \text{where } K = e^C \]

### Step 12: Solve for \( y \)
Rearranging gives:
\[ y = Kx(1+y) \]
\[ y - Kxy = Kx \]
\[ y(1 - Kx) = Kx \]
Thus,
\[ y = \frac{Kx}{1 - Kx} \]

### Step 13: General solution
The general solution can be expressed as:
\[ x + xy - Cy = 0 \quad \text{where } C = \frac{1}{K} \]

### Final Result
The general solution of the differential equation is:
\[ x + xy - Cy = 0 \]

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