Solve:

dy/dx = 1 + x + y + xy

Answer given in the book is : log | 1 + y| = x + 1/2 x2 + C

    dydx = 1+x+y +xy dydx = (1+x)(1+y) dy1+y=(1+x)dx            [variable seperable]Integrating both sides, we get dy1+y = (1+x)dx ln|1+y| =x+x22 +C

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