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I need to solve this ODE

$$ y' = \dfrac{1}{x\cos(y) + \sin(2y)}$$

Could you give me any hints? I don't even know how to start.

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    $\begingroup$ Maybe to first write this equation like $x'=...$ (to eliminate that fraction). $\endgroup$
    – Cortizol
    Commented Jun 14, 2014 at 16:25
  • $\begingroup$ @Cortizol With your hint the solution is straightforward. $\endgroup$ Commented Jun 14, 2014 at 16:44

3 Answers 3

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We have $\sin(2y)=2\sin(y)\cos(y)$, thus we can write the equation as: $$y'=\frac{1}{\cos(y)(x+2\sin(y))}$$ Now let $z=\sin(y)$, so that we obtain: $$z'=y'\cos(y)=\frac{1}{x+2z}$$ Can you take it from here?

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  • $\begingroup$ Not really sure how to take it from there. I have understood what you did, though (Is hard for me to see those changes of variable so quickly =( $\endgroup$
    – José D.
    Commented Jun 14, 2014 at 16:39
  • $\begingroup$ @Trollkemada If you want a further big hint, try substitution with $u=x+2z$. As for "seeing" substitutions, it's just a matter of training your intuition and trying around a bit. $\endgroup$ Commented Jun 14, 2014 at 16:40
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This is Ist order Linear,

$\frac{dx}{dy}-x\cos y=\sin2y $

So,

$\frac{dx}{dy}+xP(y)=Q(y)$

So, making use of general solution,

$ x = e^{-\int P(y)dy} \Big[\int Q(y)e^{\int P(y)dy}dy+C \Big]$

$ = e^{\sin y}\Big[-2e^{-\sin y}(\sin y+1) + C\Big]$

$ x = -2(\sin y +1) + C e^{\sin y}$

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Continuing after georg's answer, the solution in terms of $y$ is given by $$y=-\sin ^{-1}\Big( W\left(C e^{-(1+\frac{x}{2})}\right)+(1+\frac{x}{2})\Big)$$

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