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$y' - \frac{x}{(x^2+1)}y = 2x(x^2+1)$

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:

$y' - \frac{x}{(x^2+1)}y = 2x(x^2+1)$

$\frac{dy}{dx} = 2x(x^2 + 1) + \frac{x}{x^2+1}y$

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

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

$[2x^2 + 2 + \frac{1}{x^2+1}y] dx + [\frac{-1}{x}] dy = 0$

So that is the exact equation I got. Then:

$\frac{dM}{dy} = \frac{1}{x^2+1}$ 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?

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1 Answer 1

$$ \frac{\mathrm d}{\mathrm dx}\left(\frac{y}{\sqrt{x^2+1}}\right)=2x\sqrt{x^2+1}=\frac23\frac{\mathrm d}{\mathrm dx}\left(\sqrt{x^2+1}(x^2+1)\right) $$

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Sorry, I am not seeing what that does. –  deffq Mar 12 '13 at 13:06
    
Can you rewrite the LHS of the identity in my post as a multiple of the LHS of the ODE in your post? –  Did Mar 12 '13 at 17:41

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