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Find the particular solution of the differential equation 

\(\frac{dy}{dx}\) = 1 + x + y + xy, given that y = 0 when x = 1.

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Given:

\(\frac{dy}{dx}\) = (1 + x)(1 + y)

⇒ \(\frac{dy}{1+y}\) = (1 + x)dx

⇒ log |y + 1| = (x + \(\frac{x^2}2\) + c)

⇒ Now, for y = 0 and x = 1,

We have,

⇒ 0 = 1 + \(\frac{1}2\) + c

⇒ c = - \(\frac{3}2\)

⇒ log |y + 1| = \(\frac{x^2}2\) + x - \(\frac{3}2\)

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