# Solve $y' = \frac{1}{x\cos(y) + \sin(2y)}$

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.

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

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?

• 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 =( Commented Jun 14, 2014 at 16:39
• @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. Commented Jun 14, 2014 at 16:40

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,

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

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

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

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