The solution of the equation
`(x^(2)-xy)dy= (xy+y^(2))dx` is

Options

A. `xy=ce^(-y//x)`
B. `xy=ce^(-x//y)`
C. `yx^(2)=ce^(1//x)`
D. None of these

Answer

To solve the differential equation \((x^2 - xy) dy = (xy + y^2) dx\), we will follow a systematic approach. Here’s the step-by-step solution:

### Step 1: Rearranging the Equation
We start by rearranging the given equation to isolate \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{xy + y^2}{x^2 - xy} \]

### Step 2: Dividing by \(x^2\)
Next, we divide both the numerator and the denominator by \(x^2\) to express the equation in terms of \(\frac{y}{x}\):

\[ \frac{dy}{dx} = \frac{\frac{y}{x} + \left(\frac{y}{x}\right)^2}{1 - \frac{y}{x}} \]

Let \(v = \frac{y}{x}\). Then, we have \(y = vx\) and we can differentiate \(y\) with respect to \(x\):

\[ \frac{dy}{dx} = v + x\frac{dv}{dx} \]

### Step 3: Substituting \(y\) in Terms of \(v\)
Substituting \(y = vx\) into the equation gives:

\[ v + x\frac{dv}{dx} = \frac{v + v^2}{1 - v} \]

### Step 4: Rearranging the Equation
Now, we rearrange the equation to isolate terms involving \(v\) and \(x\):

\[ x\frac{dv}{dx} = \frac{v + v^2}{1 - v} - v \]

Combining the terms on the right-hand side:

\[ x\frac{dv}{dx} = \frac{v + v^2 - v(1 - v)}{1 - v} = \frac{v + v^2 - v + v^2}{1 - v} = \frac{2v^2}{1 - v} \]

### Step 5: Separating Variables
Now we separate the variables \(v\) and \(x\):

\[ \frac{1 - v}{2v^2} dv = \frac{1}{x} dx \]

### Step 6: Integrating Both Sides
We will integrate both sides:

\[ \int \frac{1 - v}{2v^2} dv = \int \frac{1}{x} dx \]

This can be split into two integrals:

\[ \frac{1}{2} \int \left(\frac{1}{v^2} - \frac{1}{v}\right) dv = \ln |x| + C \]

Calculating the integrals:

\[ \frac{1}{2} \left(-\frac{1}{v} - \ln |v|\right) = \ln |x| + C \]

### Step 7: Substituting Back for \(v\)
Now substituting back \(v = \frac{y}{x}\):

\[ -\frac{1}{2} \left(\frac{x}{y}\right) - \frac{1}{2} \ln \left(\frac{y}{x}\right) = \ln |x| + C \]

### Step 8: Simplifying the Expression
We can simplify this expression to get the final solution in terms of \(x\) and \(y\):

\[ \frac{x}{y} + \ln \left(\frac{y}{x}\right) = -2\ln |x| + C' \]

### Final Solution
This can be rearranged to express \(y\) in terms of \(x\):

\[ C e^{-\frac{x}{y}} = xy \]

Thus, the solution to the differential equation is:

\[ C e^{-\frac{x}{y}} = xy \]