Question

The solution of the differential equation `(1 + y^2) + (x - e^(tan^-1y)) (dy)/(dx)` = 0, is ______.

विकल्प

• `xe^(2tan^-1y) = e^(tan^-1y) + k`

• `(x - 2) = ke^(2tan^-1y`

• `2xe^(tan^-1y) = e^(2tan^-1y) + k`

• `xe^(tan^-1y) = tan^(-1y) + k`

Answer 1

The solution of the differential equation `(1 + y^2) + (x - e^(tan^-1y)) (dy)/(dx)` = 0, is `underlinebb(2xe^(tan^-1y) = e^(2tan^-1y) + k)`.

Explanation:

`(1 + y^2) + (x - e^(tan^-1y)) (dy)/(dx)` = 0

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

It is form of linear differential equation.

I.F. = `e^(int1/(1 + y^2)) = e^(tan^-1y)`

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

`x(e^(tan^-1y)) = e^((tan^-1y)^2)/2 + C`  ...`[∵ inte^(2x)dx = e^(2x)/2]`

∴ `2xe^(tan^-1y) = e^(2tan^-1y) + k`