Question

The integrating factor of the differential equation.

`(1 - y^2) dx/dy + yx = ay(-1 < y < 1)` is ______.

Options

• `1/(y^2 - 1)`

• `1/sqrt(y^2 - 1)`

• `1/(1 - y^2)`

• `1/sqrt(1 - y^2)`

Answer 1

The integrating factor of the differential equation.

`(1 - y^2) dx/dy + yx = ay(-1 < y < 1)` is `underline(1/sqrt(1 - y^2)).`

Explanation:

The differential equation is,

`(1 – y^2)dy/dx + yx = ay`

or  `dx/dy + y/(1 - y^2) x = y/(1 - y^2)`

Comparing with `dx/dy + Px = Q`,

`P = y/(1 - y^2), Q = y/(1 - y^2)`

`int P dx = int y/(1 - y^2)  dy`

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

Let `= - 1/2 int (- 2y)/(1 - y^2)  dy`

`1 - y^2` = t

∴ - 2y dy = dt

`= - 1/2 int dt/t = - 1/2 log t`

`= - 1/2 log (1 - y^2)`

`= log  1/sqrt(1 - y^2)`

`I.F. = e^(int P dx) = e^(log 1 sqrt(1 - y^2))`

`= 1/sqrt(1 - y^2)`