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Solve differential equation using integrating factor $$y'-y/x= \tan(y/x)$$

I get the integrating factor to be $1/x$

Multiply that out and the LHS $= d/dx(y (1/x))$ and the RHS then becomes $\tan(y/x) (1/x)$

Don't know where to go from that.

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    $\begingroup$ Have you considered simplifying the argument of the $\tan$: viz., set $u=y/x$, so $y'=xu'+u$, and see what happens? $\endgroup$ – Chappers Mar 8 '15 at 18:28
  • $\begingroup$ won't that make it messier? as you try and rearrange before integrating? $\endgroup$ – Jessica Mar 8 '15 at 18:52
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    $\begingroup$ "Integrating factor" is from the horror shop of mathematical misconceptions. $\endgroup$ – Christian Blatter Mar 8 '15 at 20:35
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Don't agree with 1/x as integrating factor. Following Chappers idea works perfect.

$\frac{\text{dy}}{\text{dx}}=\frac{y}{x}+\tan \left(\frac{y}{x}\right)$

Substitude:

$\frac{y}{x}=u$

$y=x u$

$\frac{\text{dy}}{\text{dx}}=x\frac{\text{du} }{\text{dx}}+u$

$ x\frac{\text{du}}{\text{dx}}+u=u+\tan (u)$

$ x\frac{\text{du}}{\text{dx}}=\tan (u)$

$\frac{\text{du}}{\tan (u)}=\frac{\text{dx}}{x}$

Now on both sides logarithmic derivatives:

$cos (u)\frac{\text{du} }{\sin (u)}=\frac{\text{dx}}{x}$

$\frac{d}{\text{du}}(\ln \sin (u))=\frac{d}{\text{dx}}(\ln (x))$

$\ln (\sin (u))=c+\ln (x)$

Solving y should be easy: $\sin (u)=C x$

$u=\arcsin (C x)$

$y=x \arcsin (C x)$

In my opinion, conditions for integrating factor are hard to solve.

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    $\begingroup$ The equation in its original form is not linear, and therefore cannot have an integrating factor anyway. $\endgroup$ – Dylan Mar 10 '15 at 2:47

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