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Question
Solve the differential equation:
`dy/dx = 1 +x+ y + xy`
Solution
`dy/dx =1 + x+ y +xy`
=(1 + x)+ y (1+x)
= (1 + x) (1 + y)
∴ `dy/(1+y) = (1+x) dx`
Integrating on both sides, we get
`intdy/(1+y) =int (1+x) dx`
∴ `log | 1+y| = x + x^2/2 + c`
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