`(dy)/(dx)=(2xy)/(x^(2)-y^(2))`

Explanation

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

### Step 1: Rewrite the Equation
We start with the given equation:
\[ \frac{dy}{dx} = \frac{2xy}{x^2 - y^2} \]

### Step 2: Separate Variables
We can rewrite the equation to separate variables:
\[ \frac{dy}{dx} = \frac{2xy}{(x^2 - y^2)} \implies \frac{dy}{y} = \frac{2x}{x^2 - y^2} dx \]

### Step 3: Substitute \(v = \frac{y}{x}\)
Let \(v = \frac{y}{x}\), then \(y = vx\). Differentiate both sides with respect to \(x\):
\[ \frac{dy}{dx} = v + x\frac{dv}{dx} \]

### Step 4: Substitute in the Equation
Substituting \(y = vx\) into the equation gives:
\[ v + x\frac{dv}{dx} = \frac{2x(vx)}{x^2 - (vx)^2} \]
This simplifies to:
\[ v + x\frac{dv}{dx} = \frac{2vx^2}{x^2(1 - v^2)} = \frac{2v}{1 - v^2} \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ x\frac{dv}{dx} = \frac{2v}{1 - v^2} - v \]
This simplifies to:
\[ x\frac{dv}{dx} = \frac{2v - v(1 - v^2)}{1 - v^2} = \frac{v(1 + v^2)}{1 - v^2} \]

### Step 6: Separate Variables Again
Rearranging yields:
\[ \frac{1 - v^2}{v(1 + v^2)} dv = \frac{dx}{x} \]

### Step 7: Integrate Both Sides
Now we integrate both sides:
\[ \int \frac{1 - v^2}{v(1 + v^2)} dv = \int \frac{dx}{x} \]

### Step 8: Simplifying the Left Side
We can break down the left side:
\[ \frac{1 - v^2}{v(1 + v^2)} = \frac{1}{v} - \frac{v}{1 + v^2} \]
Thus, we have:
\[ \int \left(\frac{1}{v} - \frac{v}{1 + v^2}\right) dv = \int \frac{dx}{x} \]

### Step 9: Perform the Integrations
Integrating gives:
\[ \ln |v| - \frac{1}{2} \ln |1 + v^2| = \ln |x| + C \]

### Step 10: Exponentiate to Solve for \(v\)
Exponentiating both sides leads to:
\[ \frac{v}{\sqrt{1 + v^2}} = kx \]
where \(k = e^C\).

### Step 11: Substitute Back for \(y\)
Recalling that \(v = \frac{y}{x}\), we substitute back:
\[ \frac{\frac{y}{x}}{\sqrt{1 + \left(\frac{y}{x}\right)^2}} = kx \]
This simplifies to:
\[ \frac{y}{\sqrt{x^2 + y^2}} = kx \]

### Step 12: Final Rearrangement
Rearranging gives us the final form:
\[ y = kx\sqrt{x^2 + y^2} \]