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