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Divide the entire equation by $(1+x^2)$ to get the standard form of a first-order linear differential equation:
Comparing the equation $\frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{4x^2}{1+x^2}$ with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we have:
The integrating factor is given by $IF = e^{\int P(x) dx}$. In this case, $P(x) = \frac{2x}{1+x^2}$. So,
To find $\int \frac{2x}{1+x^2} dx$, let $u = 1+x^2$, then $du = 2x dx$. Thus,
Now we can find the integrating factor:
The general solution of the differential equation is given by:
Substitute $IF = (1+x^2)$ and $Q(x) = \frac{4x^2}{1+x^2}$ into the general solution:
Simplify the integral:
The integral of $4x^2$ is $\frac{4}{3}x^3$. Therefore,
Divide both sides by $(1+x^2)$ to solve for $y$:
Final Answer: $y = \frac{4x^3}{3(1+x^2)} + \frac{C}{1+x^2}$