If `xy(dy)/(dx) =(1+y^(2))/(1+x^(2)) (1+x+x^(2))`, then its solution is …………….

Answer

To solve the differential equation given by

\[ xy \frac{dy}{dx} = \frac{1+y^2}{1+x^2} (1+x+x^2), \]

we will follow the steps of separation of variables and integration.

### Step 1: Separate the Variables

We start by rearranging the equation to separate the variables \(y\) and \(x\):

\[ xy \frac{dy}{dx} = \frac{(1+y^2)(1+x+x^2)}{1+x^2}. \]

We can rewrite this as:

\[ \frac{1+y^2}{y} dy = \frac{(1+x+x^2)}{x(1+x^2)} dx. \]

### Step 2: Simplify Each Side

Now, we simplify both sides:

- Left-hand side (LHS):
\[ \frac{1+y^2}{y} dy = \left(\frac{1}{y} + y\right) dy. \]

- Right-hand side (RHS):
\[ \frac{(1+x+x^2)}{x(1+x^2)} dx = \left(\frac{1}{x} + \frac{1}{1+x^2}\right) dx. \]

### Step 3: Integrate Both Sides

Now we integrate both sides:

1. **Integrating LHS**:
\[ \int \left(\frac{1}{y} + y\right) dy = \int \frac{1}{y} dy + \int y dy = \ln |y| + \frac{y^2}{2} + C_1. \]

2. **Integrating RHS**:
\[ \int \left(\frac{1}{x} + \frac{1}{1+x^2}\right) dx = \int \frac{1}{x} dx + \int \frac{1}{1+x^2} dx = \ln |x| + \tan^{-1}(x) + C_2. \]

### Step 4: Combine the Results

Equating the results from both integrations, we have:

\[ \ln |y| + \frac{y^2}{2} = \ln |x| + \tan^{-1}(x) + C, \]

where \(C = C_2 - C_1\) is a constant.

### Step 5: Final Form

To express this in a more standard form, we can rearrange it:

\[ \ln |y| = \ln |x| + \tan^{-1}(x) + C - \frac{y^2}{2}. \]

This is the implicit solution of the given differential equation.