The solution of y' - y = 1, y(0) = 1 is given by y(x) =
Answer & explanation
Correct answer: option 4
$\frac{dy}{dx}=1+y$; ln (1 + y) = x + c; y(0) = 1; y = 2ex - 1
The solution of y' - y = 1, y(0) = 1 is given by y(x) =
Correct answer: option 4
$\frac{dy}{dx}=1+y$; ln (1 + y) = x + c; y(0) = 1; y = 2ex - 1