Solve the following differential equation
`x^(2)(dy)/(dx) = x^(2) + xy - y^(2)`

Answer

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

### Step 1: Rearranging the Equation
We start by rewriting the equation to isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{x^2 + xy - y^2}{x^2} \]

### Step 2: Substituting \(y = ux\)
Next, we use the substitution \(y = ux\), where \(u\) is a function of \(x\). Then, we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = u + x \frac{du}{dx} \]

### Step 3: Substitute into the Equation
Now we substitute \(y = ux\) and \(\frac{dy}{dx}\) into the rearranged equation:
\[ u + x \frac{du}{dx} = \frac{x^2 + x(ux) - (ux)^2}{x^2} \]
This simplifies to:
\[ u + x \frac{du}{dx} = \frac{x^2 + ux^2 - u^2x^2}{x^2} \]
\[ u + x \frac{du}{dx} = 1 + u - u^2 \]

### Step 4: Simplifying the Equation
Now, we can simplify this equation:
\[ x \frac{du}{dx} = 1 - u^2 \]

### Step 5: Separating Variables
We separate the variables \(u\) and \(x\):
\[ \frac{du}{1 - u^2} = \frac{dx}{x} \]

### Step 6: Integrating Both Sides
Now we integrate both sides:
\[ \int \frac{du}{1 - u^2} = \int \frac{dx}{x} \]
The left side integrates to:
\[ \frac{1}{2} \ln \left| \frac{1 + u}{1 - u} \right| + C_1 \]
And the right side integrates to:
\[ \ln |x| + C_2 \]

### Step 7: Equating and Simplifying
Equating both sides gives us:
\[ \frac{1}{2} \ln \left| \frac{1 + u}{1 - u} \right| = \ln |x| + C \]
where \(C = C_2 - C_1\). We can exponentiate both sides to eliminate the logarithm:
\[ \left| \frac{1 + u}{1 - u} \right|^{1/2} = K |x| \quad \text{(where \(K = e^{2C}\))} \]
Squaring both sides results in:
\[ \frac{1 + u}{1 - u} = K^2 x^2 \]

### Step 8: Solving for \(u\)
Now we solve for \(u\):
\[ 1 + u = K^2 x^2 (1 - u) \]
\[ 1 + u = K^2 x^2 - K^2 x^2 u \]
\[ u + K^2 x^2 u = K^2 x^2 - 1 \]
\[ u(1 + K^2 x^2) = K^2 x^2 - 1 \]
\[ u = \frac{K^2 x^2 - 1}{1 + K^2 x^2} \]

### Step 9: Substituting Back for \(y\)
Recalling that \(y = ux\):
\[ y = x \cdot \frac{K^2 x^2 - 1}{1 + K^2 x^2} \]
Thus, the solution to the differential equation is:
\[ y = \frac{K^2 x^3 - x}{1 + K^2 x^2} \]