Answer :

Given the differential equation





Integrating both sides we have,




log|y| + log|1 – y| = log|1 + x| + logc


log|y(1 – y)| = log|c(1 + x)|


y(1 – y) = c(1 + x) ……(1)


Since, the equation passes through (2,2), So,


2(1 – 2) = c(1 + 2)


– 2 = c×3


c =


Therefore, equation (1) becomes


y(1 – y) = (1 + x)


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