The number of solutions of when y(1) = 2 is:

Integrate

⇒ log(y + 1) = log(x – 1) -log c

⇒ log(y + 1) + log c = log(x – 1)

Using log a + log b = log ab

Now it is given that y(1) = 2 which means when x = 1, y = 2

Substitute x = 1 and y = 2 in (a)

⇒ c =0

⇒ x – 1 = 0

So only one solution exists.