Integrate

⇒ log(1 + y) = x + c …(a)

Now it is given that y(0) = 1 which means when x = 0 y = 1

Hence substitute x = 0 and y = 1 in (a)

⇒ log(1 + y) = x + c

⇒ log(1 + 1) = 0 + c

⇒ c = log2

Put c = log2 back in (a)

⇒ log(1 + y) = x + log2

⇒ log(1 + y) – log2 = x

Using log a – log b = log

⇒ 1 + y = 2e^x

⇒ y = 2e^x – 1

Hence solution of differential equation is y = 2e^x – 1