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I am asked to find the integrating factor and solve.

$$ y\sin(y)dx + x(\sin(y) - y\cos(y))dy = 0.$$

I'm not sure on how to put this in the form of

$$y' + p(x)y = f(x)$$

to solve the equation. Or is there another method to use?

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You can separate dx and dy here:

$\frac {-(sin(y)−ycos(y))}{ysin(y)}dy=\frac {dx}{x}$

And solve the now separated variables equation, can you go on from here?

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  • $\begingroup$ All that would be left is integrate both sides? $\endgroup$ – cisco Feb 13 '17 at 5:27
  • $\begingroup$ Yep, easy peasy. This is the most straightforward an ode can be. $\endgroup$ – petru Feb 13 '17 at 5:50
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For an equation in the form $$Mdx +Ndy=0$$ where $M$ and $N$ are functions of $x$ and $y$ the integrating factor is $$\mu = e^{\displaystyle\int (M_y - N_x)/N dx}$$

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