`x+y(dy)/(dx) =x^(2) +y^(2) `

Answer

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

### Step 1: Substitute \( u = x^2 + y^2 \)

Let \( u = x^2 + y^2 \). Then, we can differentiate \( u \) with respect to \( x \):

\[ \frac{du}{dx} = \frac{d}{dx}(x^2 + y^2) = 2x + 2y \frac{dy}{dx} \]

### Step 2: Rearrange the equation

From the original equation, we can express \( \frac{dy}{dx} \):

\[ y \frac{dy}{dx} = x^2 + y^2 - x \]

### Step 3: Substitute \( \frac{dy}{dx} \) into the differentiated equation

Now we can substitute \( \frac{dy}{dx} \) from the rearranged equation into the expression for \( \frac{du}{dx} \):

\[ \frac{du}{dx} = 2x + 2y \frac{dy}{dx} = 2x + 2y \left( \frac{x^2 + y^2 - x}{y} \right) \]

### Step 4: Simplify the expression

This simplifies to:

\[ \frac{du}{dx} = 2x + 2(x^2 + y^2 - x) = 2x + 2u - 2x = 2u \]

### Step 5: Solve the resulting equation

Now we have a simpler differential equation:

\[ \frac{du}{dx} = 2u \]

This is a separable equation. We can separate variables:

\[ \frac{du}{u} = 2dx \]

### Step 6: Integrate both sides

Integrating both sides gives:

\[ \ln |u| = 2x + C \]

where \( C \) is the constant of integration.

### Step 7: Solve for \( u \)

Exponentiating both sides, we get:

\[ u = e^{2x + C} = e^{2x} \cdot e^C \]

Let \( k = e^C \), then:

\[ u = k e^{2x} \]

### Step 8: Substitute back for \( u \)

Recall that \( u = x^2 + y^2 \), so we have:

\[ x^2 + y^2 = k e^{2x} \]

This is the general solution of the given differential equation.

### Final Solution

The final solution to the differential equation \( x + y \frac{dy}{dx} = x^2 + y^2 \) is:

\[ x^2 + y^2 = k e^{2x} \]