Question

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

पर्याय

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

• `x  e^(tan^-1 y) = tan^-1 y + C`

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

• `(x - 2) = C  e^(-tan^-1 y)`

Answer 1

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

Explanation:

Given, differential equation is

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

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

`\implies dx/dy + x/(1 + y^2) = e^(tan^-1 y)/(1 + y^2)`, which is a linear differential equation.

Here, P = `1/(1 + y^2)`, Q = `e^(tan^-1 y)/(1 + y^2)`

IF = `e^(intPdy)`

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

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

∴ Solution of the differential equation is

x . IF = `int Q . IF  dy + C`

`xe^(tan^-1 y) = int (e^(tan^-1 y))/(1 + y^2) e^(tan^-1 y) dx + C/2`

`\implies x  e^(tan^-1 y) = (e^(2tan^-1 y))/2 + C/2`

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