Answer:
[tex]y= -\frac{1}{e^x-x+C} + 1[/tex]
Step-by-step explanation:
[tex]\frac{dy}{dx}=(y-1)^2e^x-1[/tex]
using separation of variables
[tex]\frac{dy}{(y-1)^2} = (e^x-1)dx\\\\\int{\frac{dy}{(y-1)^2}} \,=\int {(e^x-1)dx}[/tex]
now integrate
[tex]-\frac{1}{y-1} = e^x-x +C[/tex]
remember your constant of integration
[tex](y-1) * (-\frac{1}{y-1}) = (e^x-x +C)(y-1)[/tex]
[tex]\frac{-1}{e^x-x+C} = -\frac{(e^x-x +C)(y-1)}{e^x-x +C}[/tex]
[tex]y-1 = -\frac{1}{e^x-x+C}[/tex]
[tex]y= -\frac{1}{e^x-x+C} + 1[/tex]
this is your general solution, since you were not given details about a specific solution in the problem
