The solution of the differential equation `(1+y^(2)) dx = (tan^(-1) y

1.	The solution of the differential equation `(1+y^(2)) dx = (tan^(-1) y - x) dy` isA. ` x = tan ^(-1)y -1+Ce^(-tan^(-1)y)`B. ` y = tan^(-1)y+1 +Ce^(-tan^(-1)y)`C. ` x = tan^(-1)y +Ce^(-tan^(-1))y`D. None of the above
Answer» Correct Answer - a The Given differential equation is `(1+y^(2)) dx = (tan^(-1) y - x) dy` This equation can be rewritten as ` (dx)/(dy) + x/(1+y^(2)) = (tan^(-1)y)/(1+y^(2))` This is of the form , `(dx)/(dy) +Px = Q` Here, ` P = 1/(1+y^(2)) and Q = (tan^(-1)y)/(1+y^(2))` ` :. IF = e^(int 1/(1+y^(2))dy) = e^(tan-1y)` Therefore , required solution is ` xe^(tan-1y) = int e^(tan^(-1)) * (tan^(-1)y)/(1+y^(2)) dy +C` ` rArr xe^(tan-1y) = (tan ^(-1)y-1)e^(tan^(-1)y) +C` ` rArr x = tan^(-1) y - 1 +Ce ^(-tan ^(-1)y)`

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The solution of the differential equation `(1+y^(2)) dx = (tan^(-1) y - x) dy` isA. ` x = tan ^(-1)y -1+Ce^(-tan^(-1)y)`B. ` y = tan^(-1)y+1 +Ce^(-tan^(-1)y)`C. ` x = tan^(-1)y +Ce^(-tan^(-1))y`D. None of the above

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