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I have the following DE:

$ xy' = y + x\cos^2\left(\frac{y}{x}\right) $

I then rule out the possible methods of solving it:

  • Not separable
  • Not homogeneous
  • Not exact
  • Possible integrating factor in $x$? No.
  • Possible integrating factor in $y$? No.
  • Not linear

Above are the only ways I have learned to solve DEs. To help, I've rewritten the DE in this form:

$ M(x,y)dx + N(x,y)dy = 0 $ $ \left(y + x\cos^2\left(\frac{y}{x}\right)\right)dx - xdy =0 $ $ M_{y} = 1 - 2\cos\left(\frac{y}{x}\right)\sin\left(\frac{y}{x}\right) $ $ N_{x} = -1 $

I'm completely lost now. I can't seem to find any integrating factors (the $\frac{y}{x}$) inside the trigs are making it so that I can't get things only in terms of $x$ or $y$.

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    This is the right answer unless I made a mistake in my calculations. $y=x\arctan(kln(x))$ k=const.2012-10-07

1 Answers 1

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Dividing both sides by $x$, we get $\tag{1}y' =\frac{y}{x} + \cos^2\left(\frac{y}{x}\right).$ If we let $u=\displaystyle\frac{y}{x}$, then we have $y=ux$ and $\frac{dy}{dx}=x\frac{du}{dx}+u.$ Substitute this into $(1)$, we have $x\frac{du}{dx}+u=u+\cos^2(u).$ I think you can solve it starting from here.