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

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:

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

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

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

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

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

So that is the exact equation I got. Then:

$$\frac{\partial M}{\partial y} = \frac{1}{x^2+1} \text{ and } \frac{\partial N}{\partial x} = \frac{1}{x^2}$$

And now I'm stuck. I tried to get an integrating factor here to make $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, 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 $\frac{\partial M}{\partial y}$ and $\frac{\partial N}{\partial x}$ 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?


1 Answer 1


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

  • $\begingroup$ Sorry, I am not seeing what that does. $\endgroup$
    – deffq
    Commented Mar 12, 2013 at 13:06
  • $\begingroup$ Can you rewrite the LHS of the identity in my post as a multiple of the LHS of the ODE in your post? $\endgroup$
    – Did
    Commented Mar 12, 2013 at 17:41

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