`x^(2)dy+y(x+y)dx=0`

Explanation

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

### Step 1: Rewrite the equation
We start by rewriting the given equation in a more manageable form:
\[ x^2 dy + y(x+y)dx = 0 \]
This can be rearranged to express \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -\frac{y(x+y)}{x^2} \]

### Step 2: Check for homogeneity
We need to check if the function is homogeneous. A function \( f(x, y) \) is homogeneous of degree \( n \) if:
\[ f(\lambda x, \lambda y) = \lambda^n f(x, y) \]
Substituting \( \lambda x \) and \( \lambda y \) into our function:
\[ f(\lambda x, \lambda y) = -\frac{\lambda y(\lambda x + \lambda y)}{\lambda^2 x^2} = -\frac{\lambda^2 y(x+y)}{\lambda^2 x^2} = -\frac{y(x+y)}{x^2} \]
This shows that the function is homogeneous of degree 0.

### Step 3: Substitute \( y = vx \)
Let \( y = vx \), where \( v \) is a function of \( x \). Then, we differentiate:
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]
Substituting \( y = vx \) into the equation gives:
\[ v + x \frac{dv}{dx} = -\frac{vx(x + vx)}{x^2} \]
This simplifies to:
\[ v + x \frac{dv}{dx} = -v(v + 1) \]

### Step 4: Rearranging the equation
Rearranging the equation leads to:
\[ x \frac{dv}{dx} = -v(v + 1) - v = -v(v + 2) \]
This can be rewritten as:
\[ \frac{dv}{v(v + 2)} = -\frac{dx}{x} \]

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

### Step 6: Simplifying the equation
This can be simplified to:
\[ \ln \left| \frac{v}{v + 2} \right| = -2 \ln |x| + C' \]
Exponentiating both sides results in:
\[ \frac{v}{v + 2} = \frac{C}{x^2} \]
Substituting back \( v = \frac{y}{x} \):
\[ \frac{\frac{y}{x}}{\frac{y}{x} + 2} = \frac{C}{x^2} \]
Cross-multiplying gives:
\[ y = \frac{C x^2}{x + 2C} \]

### Final Solution
Thus, the final solution to the differential equation is:
\[ x^2 y = C(y + 2x) \]