Find the particular solution of the differential equation
`2xy+y^(2)-2x^(2)(dy)/(dx)=0`, it being given that `y=2` when `x=1`.

Explanation

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

### Step 1: Rearranging the Equation
Start by isolating \(\frac{dy}{dx}\):
\[ 2x^2 \frac{dy}{dx} = 2xy + y^2 \]
\[ \frac{dy}{dx} = \frac{2xy + y^2}{2x^2} \]

### Step 2: Identifying Homogeneity
The equation is homogeneous because if we replace \(x\) and \(y\) with \(\lambda x\) and \(\lambda y\), the equation remains unchanged. This allows us to use the substitution \(y = vx\), where \(v\) is a function of \(x\).

### Step 3: Substituting \(y = vx\)
Substituting \(y = vx\) into the equation gives:
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]
Substituting this into the rearranged equation:
\[ v + x \frac{dv}{dx} = \frac{2x(vx) + (vx)^2}{2x^2} \]
\[ v + x \frac{dv}{dx} = \frac{2v + v^2}{2} \]

### Step 4: Simplifying the Equation
Multiply through by \(2\) to eliminate the fraction:
\[ 2v + 2x \frac{dv}{dx} = 2v + v^2 \]
Cancel \(2v\) from both sides:
\[ 2x \frac{dv}{dx} = v^2 \]

### Step 5: Separating Variables
Rearranging gives:
\[ \frac{dv}{v^2} = \frac{1}{2x} dx \]

### Step 6: Integrating Both Sides
Integrate both sides:
\[ \int \frac{dv}{v^2} = \int \frac{1}{2x} dx \]
This results in:
\[ -\frac{1}{v} = \frac{1}{2} \ln |x| + C \]

### Step 7: Substituting Back for \(v\)
Recall that \(v = \frac{y}{x}\), so substituting back gives:
\[ -\frac{x}{y} = \frac{1}{2} \ln |x| + C \]

### Step 8: Applying the Initial Condition
Using the initial condition \(y = 2\) when \(x = 1\):
\[ -\frac{1}{2} = \frac{1}{2} \ln |1| + C \]
Since \(\ln |1| = 0\):
\[ C = -\frac{1}{2} \]

### Step 9: Finalizing the Solution
Substituting \(C\) back into the equation gives:
\[ -\frac{x}{y} = \frac{1}{2} \ln |x| - \frac{1}{2} \]
Multiplying through by \(-2\) yields:
\[ \frac{2x}{y} = -\ln |x| + 1 \]
Rearranging gives the final form:
\[ \ln |x| + \frac{2x}{y} = 1 \]

### Final Answer
Thus, the particular solution to the differential equation is:
\[ \ln |x| + \frac{2x}{y} = 1 \]