The solution of y' - y = 1, y(0) = 1 is given by y(x) =

-exp .(x) -exp.(-x) -1 2 exp.(x)-1

2 exp.(x)-1

$\frac{dy}{dx}=1+y$; ln (1 + y) = x + c; y(0) = 1; y = 2ex - 1