Solution of the diff. equation
`x^(2)dy+y(x+y)dx=0` is

Options

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

Answer

To solve the differential equation \( x^2 dy + y(x+y)dx = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the given differential equation:
\[ x^2 dy = -y(x+y) dx \]
Dividing both sides by \( x^2 y \) (assuming \( y \neq 0 \)):
\[ \frac{dy}{y} = -\frac{x+y}{x^2} dx \]

### Step 2: Expressing in Terms of \( \frac{dy}{dx} \)
Now we can express \( \frac{dy}{dx} \) in terms of \( y \) and \( x \):
\[ \frac{dy}{dx} = -\frac{y(x+y)}{x^2} \]

### Step 3: Substituting \( v = \frac{y}{x} \)
Let \( y = vx \), where \( v \) is a function of \( x \). Then, we differentiate:
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]
Substituting this into our equation gives:
\[ v + x \frac{dv}{dx} = -\frac{vx + vx^2}{x^2} \]
This simplifies to:
\[ v + x \frac{dv}{dx} = -\frac{v}{x} - v \]

### Step 4: Rearranging and Separating Variables
Rearranging the equation:
\[ x \frac{dv}{dx} = -\frac{2v}{x} \]
This can be rewritten as:
\[ \frac{dv}{v(2 + v)} = -\frac{dx}{x} \]

### Step 5: Integrating Both Sides
Now we integrate both sides:
\[ \int \frac{dv}{v(2 + v)} = \int -\frac{dx}{x} \]
Using partial fractions on the left side:
\[ \frac{1}{v(2+v)} = \frac{A}{v} + \frac{B}{2+v} \]
Solving for \( A \) and \( B \) gives:
\[ A(2+v) + Bv = 1 \]
Setting \( v = 0 \) gives \( A = \frac{1}{2} \) and setting \( v = -2 \) gives \( B = -\frac{1}{2} \):
\[ \int \left( \frac{1/2}{v} - \frac{1/2}{2+v} \right) dv = \int -\frac{dx}{x} \]
This leads to:
\[ \frac{1}{2} \ln |v| - \frac{1}{2} \ln |2+v| = -\ln |x| + C \]

### Step 6: Simplifying the Result
Combining the logarithms:
\[ \ln \left| \frac{v}{2+v} \right| = -2 \ln |x| + C' \]
Exponentiating both sides gives:
\[ \frac{v}{2+v} = \frac{C}{x^2} \]
Substituting back \( v = \frac{y}{x} \):
\[ \frac{\frac{y}{x}}{2 + \frac{y}{x}} = \frac{C}{x^2} \]
This simplifies to:
\[ \frac{y}{y + 2x} = \frac{C}{x^2} \]
Cross-multiplying gives:
\[ y = \frac{C x^2}{y + 2x} \]

### Final Result
The solution of the differential equation is:
\[ y + 2x = \frac{C x^2}{y} \]