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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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Answer» 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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