Differential equations?

Question

(6x-3y+2)dx - (2x-y-1)dy=0

x^4 y' = -x^3 y - csc(xy)

[1+ y tan(xy) ]dx + x tan(xy) dy = 0


thanks

Answer 1

For the first equation,

let u = 2x - y - 1; then y = 2x - u - 1, dy = 2dx - du

For both of the other equations,

let u = xy; then y = u/x, y' = u'/x - u/x², dy = (1/x)du - (u/x²)dx