$$y' - \frac{x}{(x^2+1)}y = 2x(x^2+1)$$
I need to solve this differential equation using the normal integrating factor method for 1st order linear DEs, and a second method chosen from: separable equations, homogenous equations, Bernoulli equations and exact equations. I had no problem solving it using the normal method for linear equations, but I don't see how any of these cases apply to the DE above. The only method that seems plausible is exact equations. I tried to use an integrating factor to turn it into an exact equation, but it did not work out.
This is what I've tried in terms of exact equations:
$$y' - \frac{x}{(x^2+1)}y = 2x(x^2+1)$$
$$\frac{dy}{dx} = 2x(x^2 + 1) + \frac{x}{x^2+1}y$$
$$\frac{dy}{dx} = x \left[2x^2 + 2 + \frac{1}{x^2+1}y\right]$$
$$[\frac{1}{x}] \, dy = \left[2x^2 + 2 + \frac{1}{x^2+1}y\right] \, dx$$
$$ \left[2x^2 + 2 + \frac{1}{x^2+1}\right] \, dx + \left[\frac{-1}{x}\right] \, dy = 0$$
So that is the exact equation I got. Then:
$$\frac{dM}{dy} = \frac{1}{x^2+1} \text{ and } \frac{dN}{dx} = \frac{1}{x^2}$$
And now I'm stuck. I tried to get an integrating factor here to make $\frac{dM}{dy} = \frac{dN}{dx}$, but it gets extremely complicated and it can't be what the professor intended for us to do. I think I just screwed up somewhere because $\frac{dM}{dy}$ and $\frac{dN}{dx}$ are so similar that it seems like it is the correct method to use.
I've been working on this for a long time. Can anyone here please give me a hint or tell me if I'm overlooking something glaringly obvious?
