Yes, I can do that from there. Thank you!
Just so you can check your result, I would write:
[MATH]v_1(x)=-\frac{1}{w^2}\int \sin(wx)\tan(wx)w\,dx=\frac{1}{w^2}\int \cos(wx)-\sec(wx)w\,dx=\frac{\sin(wx)-\ln(\sec(wx)+\tan(wx))}{w^2}[/MATH]
And so our particular solution is:
[MATH]y_p(x)=\left(\frac{\sin(wx)-\ln(\sec(wx)+\tan(wx))}{w^2}\right)\cos(wx)+\left(-\frac{1}{w^2}\cos(wx)\right)\sin(wx)=-\frac{1}{w^2}\cos(wx)\ln(\sec(wx)+\tan(wx))[/MATH]
And so, the general solution to the given ODE is:
[MATH]y(x)=y_h(x)+y_p(x)=c_1\cos(wx)+c_2\sin(wx)-\frac{1}{w^2}\cos(wx)\ln(\sec(wx)+\tan(wx))[/MATH]
And that would be our solution if I hadn't made a silly oversight...we have \(g(x)=\tan(qx)\)...and so we need to go all the way back to this system:
[MATH]\cos(wx)v_1'+\sin(wx)v_2'=0[/MATH]
[MATH]-w\sin(wx)v_1'+w\cos(wx)v_2'=\tan(qx)[/MATH]
As before, the first equation implies:
[MATH]v_1'=-\tan(wx)v_2'[/MATH]
And so the second equation becomes:
[MATH]-w\sin(wx)\left(-\tan(wx)v_2'\right)+w\cos(wx)v_2'=\tan(qx)[/MATH]
[MATH]w\sin^2(wx)v_2'+w\cos^2(wx)v_2'=\tan(qx)\cos(wx)[/MATH]
[MATH]v_2'=\frac{1}{w}\tan(qx)\cos(wx)[/MATH]
And I have no idea how to integrate that...W|A indicates that it involves the hypergeometric function.