(2x+y+1)y' = 1
Although this intially appeared to be a simple problem I've tried every method I know how to do on this problem and I just can't get a solution.
alternatively
(2x+y+1)y' = 1
can be written
(2x+y+1)dy = dx
-dx + (2x+y+1)dy = 0
can be made exact with the integrating factor e^(-2y)
-e^(-2y)dx + e^(-2y)(2x+y+1)dy = 0
dF/dx = -e^(-2y)
F = -xe^(-2y) + f(y)
dF/dy = 2xe^(-2y) + f ' (y) = e^(-2y)(2x+y+1)
f ' (y) = (y+1)e^(-2y)
f(y) = -e^(-2y)(y+1)
F = -xe^(-2y) -e^(-2y)(y+1) = C
-x = Ce^(2y) +(y+1)
x = -(y+1) + Ce^(2y)
as in the previous post