What is the general solution of the differential equation x dy -ydx =`y^(2)` ?

Options

A. x=cy
B. `y^(2)=cx`
C. x+xy -cy =0
D. None of these

Explanation

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

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

### Step 2: Divide by \( y^2 \)
Next, we divide both sides by \( y^2 \):
\[ \frac{x \, dy}{y^2} = \frac{y \, dx}{y^2} + 1 \]
This simplifies to:
\[ \frac{x \, dy}{y^2} = \frac{dx}{y} + 1 \]

### Step 3: Introduce a substitution
Let \( v = \frac{x}{y} \). Then, we have \( y = \frac{x}{v} \) and differentiating gives:
\[ dy = \frac{1}{v} \, dx - \frac{x}{v^2} \, dv \]

### Step 4: Substitute into the equation
Substituting \( y \) and \( dy \) into the equation, we get:
\[ \frac{x \left( \frac{1}{v} \, dx - \frac{x}{v^2} \, dv \right)}{\left( \frac{x}{v} \right)^2} = \frac{dx}{\frac{x}{v}} + 1 \]
This simplifies to:
\[ \frac{v^2}{x} \left( \frac{x}{v} \, dx - \frac{x^2}{v^2} \, dv \right) = v \, dx + 1 \]
This leads to:
\[ dx - \frac{x}{v} \, dv = v \, dx + 1 \]

### Step 5: Rearranging the terms
Rearranging gives:
\[ dx - v \, dx = \frac{x}{v} \, dv + 1 \]
Factoring out \( dx \):
\[ (1 - v) \, dx = \frac{x}{v} \, dv + 1 \]

### Step 6: Separate variables
We can separate variables:
\[ \frac{dx}{\frac{x}{v} + 1} = \frac{dv}{1 - v} \]

### Step 7: Integrate both sides
Integrating both sides:
\[ \int \frac{dx}{\frac{x}{v} + 1} = \int \frac{dv}{1 - v} \]
The left side integrates to:
\[ v \ln\left(\frac{x}{v} + 1\right) \]
And the right side integrates to:
\[ -\ln|1 - v| + C \]

### Step 8: Solve for \( y \)
After integrating and simplifying, we can express \( y \) in terms of \( x \) and \( v \):
\[ x + xy - C = 0 \]
Thus, the general solution can be expressed as:
\[ x + xy - C = 0 \]

### Final Answer
The general solution of the differential equation is:
\[ x + xy = C \]

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