Question

Solve the differential equation `(x^2 - 1) "dy"/"dx" + 2xy = 1/(x^2 - 1)`.

Answer 1

Given differential equation is `(x^2 - 1) "dy"/"dx" + 2xy = 1/(x^2 - 1)`.

Dividing by (x² – 1), we get

`"dy"/"dx" + "xy"/(x^2 - 1) = 1/(x^2 - 1)`

It is a linear differential equation of first order and first degree.

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

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

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

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

= `(x^2 - 1)`

∴ Solution of the equation is `y xx "I"."F". = int "Q" . "I"."F".  "d"x + "C"`

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

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

⇒ `y(x^2 - 1) = 1/2 log|(x - 1)/(x + 1)| + "C"`.

Hence the required solution is  `y(x^2 - 1) = 1/2 log|(x - 1)/(x + 1)| + "C"`.