Question

"(1+x)*y'-2y=0"

Answer 1

-2 y[x] + y'[x] + x y'[x] = 0

Answer 2

(1+x)*y'-2y=0
(1+x) dy/dx-2y=0
(1+x) dy/dx=2y
separating the variables
dy/y=2 dx/(1+x)
integrating
logy=2 log (1+x)+log c'
logy=log(1+x)^2+log c'
logy=log (1+x)^2 *(c')
or y=c' (1+x)^2