` xy (dy)/(dx) = x^(2) + 2y^(2) , y(1) = 0`

Answer

To solve the differential equation \( xy \frac{dy}{dx} = x^2 + 2y^2 \) with the initial condition \( y(1) = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ xy \frac{dy}{dx} = x^2 + 2y^2 \]
We can rearrange this to isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{x^2 + 2y^2}{xy} \]

### Step 2: Separate variables
Next, we can separate the variables \(y\) and \(x\):
\[ \frac{dy}{dx} = \frac{x}{y} + \frac{2y}{x} \]
This can be rewritten as:
\[ \frac{dy}{dx} - \frac{2y}{x} = \frac{x}{y} \]

### Step 3: Identify the type of differential equation
This is a first-order linear differential equation in the standard form:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
where \(P(x) = -\frac{2}{x}\) and \(Q(x) = \frac{x}{y}\).

### Step 4: Find an integrating factor
To solve this, we need to find an integrating factor \( \mu(x) \):
\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int -\frac{2}{x} \, dx} = e^{-2 \ln |x|} = \frac{1}{x^2} \]

### Step 5: Multiply through by the integrating factor
Now we multiply the entire differential equation by the integrating factor:
\[ \frac{1}{x^2} \frac{dy}{dx} - \frac{2y}{x^3} = \frac{1}{x} \]

### Step 6: Rewrite the left side
The left side can be rewritten as:
\[ \frac{d}{dx}\left(\frac{y}{x^2}\right) = \frac{1}{x} \]

### Step 7: Integrate both sides
Now we integrate both sides:
\[ \int \frac{d}{dx}\left(\frac{y}{x^2}\right) dx = \int \frac{1}{x} dx \]
This gives us:
\[ \frac{y}{x^2} = \ln |x| + C \]

### Step 8: Solve for \(y\)
Now we solve for \(y\):
\[ y = x^2(\ln |x| + C) \]

### Step 9: Apply the initial condition
We apply the initial condition \(y(1) = 0\):
\[ 0 = 1^2(\ln |1| + C) \implies 0 = 0 + C \implies C = 0 \]
Thus, the solution simplifies to:
\[ y = x^2 \ln |x| \]

### Final Solution
The final solution to the differential equation is:
\[ y = x^2 \ln |x| \]