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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