Answer :
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.
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