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
\(\Rightarrow c = -\frac{2}{3}\)
Therefore, equation (1) becomes
y(1 – y) = \(-\frac{2}{3}\)(1 + x)