Solve the Differential Equation (1-cos(x))dy + (2ysin(x) - tan(x)) dx = 0 I have tried this two ways so far and either way that I do it does not look correst. First way: As a seperable DE. (1-cos(x))dy/dx = 2ysin(x) - tan(x) simplifying I eventually get...... dx - sec(x)dx = [cscx(cosx + 1)/2y]dy and more simplifing [1- sec(x)]dx/cscx(cosx + 1) = dy/2y Then I integrated the right side and am having trouble integrating the left side. int.[tan(x) * (cosx - 1)/(cosx + 1)] = (1/2) ln (y ) +c is this a reasonable place to keep going from? if so, any tips on the integration? The second way I did this is as a first order linear DE. I got it in the form (dy/dx) + p(x)y = q(x) I used e and raised it to the integral of p(x) and did some more work y[e^[-2ln(1 - cosx)] = integrate.[ -tanx( 1 / (1-cosx))(e^[-2ln(1 - cosx))] the right side does not look like other problems I do at the end when I do the final integration. any thoughts on this...? Even make any sense?