>>11225505Let v = y', so that v' + v tan(x) = sin(2x).
Now, simply solve by finding the integrating factor. We want f(x) such that (f(x)v)' = f(x)v' + v f'(x) is something multiplied by v' + v tan(x). Well, we must be multiplying by f(x) since we need to turn v' into f(x) v'... so then we need f'(x) = f(x) tan(x) . what function has a derivative which is itself multiplied by tangent? secant. so f(x) = sec(x) and we get (sec(x)v)' = sec(x)sin(2x) = 2sin(x)cos(x)sec(x) = 2sin(x), so integrating we find sec(x)v = -2cos(x) + C, so v = -2cos^2(x) + C cos(x).
Now y' = v, so integrate once more to get the answer. I don't remember the integrals of squares of trig functions so do it yourself (via double angle). don't forget your second constant.
alternatively, just use the normal integrating factor approach you learned, it works but i dont remember it, i just find the factor on the fly.
