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Question
Solve : `"dy"/"dx" = 1 - "xy" + "y" - "x"`
Solution
`"dy"/"dx" = 1 - "xy" + "y" - "x"`
`"dy"/"dx" = (1 + "y") - "x" (1 + "y")`
`"dy"/"dx" = (1 + "y") (1 - "x")`
`"dy"/(1 + "y") = (1 - "x")"dx"`
Integrating bothe sides, we obtain
`int"dy"/(1 + "y") = int (1 - "x")"dx"`
log |1 + y| `= "x" - "x"^2/2 + "C"`
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