Question

The general solution of the differential equation (1 + y²) + `(x - e^(tan^-1 y)) "dy"/"dx"` = 0 is.

पर्याय

• `x * "e"^(tan^-1 y) = ("e"^(tan^-1 y))^2/2` + C

• `"e"^(tan^-1 y) = ("e"^(tan^-1 x))^2` + C

• `x * "e"^(tan^-1 y) = ("e"^(tan^-1) x)^2/2` + C

• `"e"^(tan^-1 y) = ("e"^(tan^-1 y))^2` + C

Answer 1

`x * "e"^(tan^-1 y) = ("e"^(tan^-1 y))^2/2` + C

Explanation:

Given equation,

(1 + y²) + `(x - e^(tan^-1 y)) "dy"/"dx"` = 0

`=> "dx"/"dy" + x/(1 + y^2) = ("e"^(tan^-1 y))/(1 + y^2)`

∴ IF = `"e"^(int 1/(1 + y^2)"dy")`

`= "e"^(tan^-1 "y")`

Solution of given differential equation is,

`x"e"^(tan^-1 "y") = int ("e"^(2 tan^-1 "y"))/(1 + y^2)`dy + c

`x "e"^(tan^-1 "y") = ("e"^(2 tan^-1 "y"))/2` + c = `("e"^(tan^-1 "y"))^2/2` + c