Question

The solution of the differential equation `("d"y)/("d"x) + (2xy)/(1 + x^2) = 1/(1 + x^2)^2` is ______.

Options

• y(1 + x²) = c + tan^–1x

• `y/(1 + x^2) = "c" + tan^-1x`

• y log(1 + x²) = c + tan^–1x

• y(1 + x²) = c + sin^–1x

Answer 1

The solution of the differential equation `("d"y)/("d"x) + (2xy)/(1 + x^2) = 1/(1 + x^2)^2` is y(1 + x²) = c + tan^–1x.

Explanation:

The given differential equation is `("d"y)/("d"x) + (2xy)/(1 + x^2) = 1/(1 + x^2)^2`

Since, it is a linear differential equation

P = `(2x)/(1 + x^2)` and Q = `1/(1 + x^2)^2`

Integrating factor I.F. = `"e"^(int Pdx)`

= `"e"^(int (2x)/(1 + x^2) "d"x)`

= `"e"^(log(1 + x^2))`

= `(1 + x^2)`

∴ Solution is `y xx "I"."F". = int "Q" xx "I"."F".  "d"x + "c"`

⇒ `y(1 + x^2) = int 1/(1 + x^2)^2 xx (1 + x^2)"d"x + "c"`

⇒ `y(1 + x^2) = int 1/((1 + x^2)) "d"x + "c"`

⇒ `y(1 + x^2) = tan^-1x + "c"`.