Solve the following differential equations :
`(dy)/(dx)-(2x)/(1+x^(2))y=x^(2)+2`.

Answer

To solve the differential equation

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

we can follow these steps:

### Step 1: Identify the form of the equation
The given equation is in the standard form of a linear first-order differential equation:

\[ \frac{dy}{dx} + P(x)y = Q(x), \]

where \( P(x) = -\frac{2x}{1+x^2} \) and \( Q(x) = x^2 + 2 \).

### Step 2: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int -\frac{2x}{1+x^2} \, dx}. \]

To compute the integral, we can use the substitution \( t = 1 + x^2 \), then \( dt = 2x \, dx \), which gives us:

\[ \int -\frac{2x}{1+x^2} \, dx = -\int \frac{dt}{t} = -\ln|t| + C = -\ln|1+x^2| + C. \]

Thus, the integrating factor becomes:

\[ \mu(x) = e^{-\ln(1+x^2)} = \frac{1}{1+x^2}. \]

### Step 3: Multiply the entire equation by the integrating factor
Now we multiply the entire differential equation by \( \mu(x) \):

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

This simplifies to:

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

### Step 4: Integrate both sides
Now we integrate both sides with respect to \( x \):

\[ \int \frac{d}{dx}\left(\frac{y}{1+x^2}\right) \, dx = \int \frac{x^2 + 2}{1+x^2} \, dx. \]

The left side simplifies to:

\[ \frac{y}{1+x^2} = \int \left(1 + \frac{1}{1+x^2}\right) \, dx = x + \tan^{-1}(x) + C. \]

### Step 5: Solve for \( y \)
Now, we multiply both sides by \( 1+x^2 \):

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

### Final Solution
Thus, the solution to the differential equation is:

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