# ​ Find the particular solution of the differential equation  dy/dx = 1 + x + y + xy, given that y = 0 when x = 1. ​

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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 - $\frac{3}2$