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

Options

A. `log x =log(x-y) +(y)/(x)+c`
B. `log x =2 log (x-y) +(y)/(x) +c`
C. `log x =log(x-y) +(x)/(y) +c`
D. None of these

Answer

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

### Step 1: Rearranging the Equation
We start by rewriting the given equation in the standard form:
\[ \frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy} \]

### Step 2: Dividing by \(x^2\)
Next, we divide both the numerator and the denominator by \(x^2\):
\[ \frac{dy}{dx} = \frac{1 + \left(\frac{y}{x}\right)^2}{1 + \frac{y}{x}} \]
Let \(v = \frac{y}{x}\), which implies \(y = vx\).

### Step 3: Differentiating \(y\)
Differentiating \(y\) with respect to \(x\) gives:
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]

### Step 4: Substituting into the Equation
Now substitute \(y = vx\) and \(\frac{dy}{dx}\) into the rearranged equation:
\[ v + x \frac{dv}{dx} = \frac{1 + v^2}{1 + v} \]

### Step 5: Rearranging Terms
Rearranging gives:
\[ x \frac{dv}{dx} = \frac{1 + v^2}{1 + v} - v \]
Simplifying the right-hand side:
\[ x \frac{dv}{dx} = \frac{1 + v^2 - v(1 + v)}{1 + v} = \frac{1 + v^2 - v - v^2}{1 + v} = \frac{1 - v}{1 + v} \]

### Step 6: Separating Variables
Now we separate the variables:
\[ \frac{1 + v}{1 - v} dv = \frac{dx}{x} \]

### Step 7: Integrating Both Sides
Integrating both sides:
\[ \int \frac{1 + v}{1 - v} dv = \int \frac{dx}{x} \]
The left side can be split:
\[ \int \left(-1 + \frac{2}{1 - v}\right) dv = \ln |x| + C \]
Thus, we have:
\[ -v + 2 \ln |1 - v| = \ln |x| + C \]

### Step 8: Back Substituting for \(v\)
Substituting back \(v = \frac{y}{x}\):
\[ -\frac{y}{x} + 2 \ln \left|1 - \frac{y}{x}\right| = \ln |x| + C \]

### Step 9: Rearranging the Equation
Rearranging gives:
\[ 2 \ln \left|1 - \frac{y}{x}\right| = \ln |x| + C + \frac{y}{x} \]

### Step 10: Exponentiating
Exponentiating both sides to eliminate the logarithm:
\[ \left|1 - \frac{y}{x}\right|^2 = K \cdot |x| e^{\frac{y}{x}} \]
where \(K = e^C\).

### Final Form
Thus, the solution to the differential equation is:
\[ \left|1 - \frac{y}{x}\right|^2 = K \cdot |x| e^{\frac{y}{x}} \]