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

Explanation

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

### Step 1: Rearranging the Equation
We start by rewriting the equation in the standard form:
\[ 2xy \, dx + (x^2 + 2y^2) \, dy = 0 \]
This can be rearranged to:
\[ \frac{dy}{dx} = -\frac{2xy}{x^2 + 2y^2} \]

### Step 2: Introducing a Substitution
Next, we introduce a substitution to simplify the equation. Let:
\[ v = \frac{y}{x} \quad \Rightarrow \quad y = vx \]
Differentiating \( y \) with respect to \( x \) gives:
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]

### Step 3: Substituting into the Equation
Substituting \( y = vx \) and \( \frac{dy}{dx} = v + x \frac{dv}{dx} \) into the rearranged equation:
\[ v + x \frac{dv}{dx} = -\frac{2x(v)}{x^2 + 2(vx)^2} \]
This simplifies to:
\[ v + x \frac{dv}{dx} = -\frac{2v}{x + 2v^2} \]

### Step 4: Isolating the Derivative
Rearranging gives:
\[ x \frac{dv}{dx} = -\frac{2v}{x + 2v^2} - v \]
Combining terms:
\[ x \frac{dv}{dx} = -\frac{2v + v(x + 2v^2)}{x + 2v^2} = -\frac{(2 + x)v + 2v^3}{x + 2v^2} \]

### Step 5: Separating Variables
Now we separate the variables:
\[ \frac{(x + 2v^2)}{(2 + x)v + 2v^3} dv = -\frac{dx}{x} \]

### Step 6: Integrating Both Sides
Integrating both sides:
\[ \int \frac{(x + 2v^2)}{(2 + x)v + 2v^3} dv = -\int \frac{dx}{x} \]

### Step 7: Solving the Integrals
After performing the integration, we will have:
\[ \log |x| = -\frac{1}{3} \log |3v + 2v^3| + C \]
This implies:
\[ 3v + 2v^3 = \frac{C}{x^3} \]

### Step 8: Substituting Back
Substituting back \( v = \frac{y}{x} \):
\[ 3\frac{y}{x} + 2\left(\frac{y}{x}\right)^3 = \frac{C}{x^3} \]
Multiplying through by \( x^3 \):
\[ 3y x^2 + 2y^3 = C \]

### Final Answer
Thus, the final solution to the differential equation is:
\[ 3yx^2 + 2y^3 = C \]