solve the ode $x^2y^\prime-1=\cos 2y$

First I wrote $y=y_h+y_p$. $y_h$ can be found easily i think: $$x^2y^\prime-1=0 \\y^\prime=\frac 1 x^2 \\ y=-\frac 1 x +c_1$$ but about finding the private solution i dont know from where to begin. suppose it was $\cos 2x$ i could have guessed $$y_p=a\cos 2x+ b\sin 2x$$ but here I can't do so. how can I find $y_p$?

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    $\begingroup$ try solving $$ \int\frac{\mathrm{dy}}{1+\cos(2y)} = \int\frac{1}{x^{2}}\mathrm{dx}$$. With also using some trig identities. $\endgroup$ – Chinny84 Jan 22 '14 at 18:16
  • $\begingroup$ This is a nonlinear equation (because of the $\cos(2\,y)$ term.) The method you are trying to use is for linear equations. $\endgroup$ – Julián Aguirre Jan 22 '14 at 18:20
  • $\begingroup$ @JuliánAguirre yes the $\cos(2y)$ term is non-linear but the equation is a first order separable ode so I can split out the y terms and x terms and integrate without a problem. If the rhs was $x\cos(2y)$ for example, then I would not be able to separate. $\endgroup$ – Chinny84 Jan 22 '14 at 18:26
  • $\begingroup$ @Chinny84 Of course. Since you explained that in your first comment, I did not see the need to repeat it. I just wanted to make sure the OP understands that the ansatz $y=y_h+y_p$ does not work in this case. $\endgroup$ – Julián Aguirre Jan 22 '14 at 18:30
  • $\begingroup$ @JuliánAguirre Oh ok. Sorry, I misread your comment. $\endgroup$ – Chinny84 Jan 22 '14 at 19:33

I think one of the best way for solving this non-linear one is to do that in light of @Chinny84. We have:

$$x^2y^\prime-1=\cos 2y\longrightarrow x^2dy=\big(1+\cos(2y)\big)dx=2\cos^2(y)dx$$ so we get $$\frac{dy}{2\cos^2(y)}=\frac{dx}{x^2}\longrightarrow 0.5\tan(y)=-\frac{1}x+C, x\neq 0$$


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