Solve the differential equation: $\displaystyle dy/dx = tan[(y-x)/x]+1 + (y/x$) for y.

Solve $\displaystyle dy/dx + (6/5)*cos[x]^2 = y^(-2/3)*e^(-x)*cos[2x]$ for y.

The first one I simplified it down to $\displaystyle Ln(x) + c = 1/4(2v - sin[2-2v] - 2)$ where $\displaystyle v = y/x$ I was wondering if there is an easier way to do this.

I have no clue on the second one O.O