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Question
`"If y" = (sec^-1 "x")^2 , "x" > 0 "show that" "x"^2 ("x"^2 - 1) (d^2"y")/(d"x"^2) + (2"x"^3 - "x") (d"y")/(d"x") - 2 = 0`
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Solution
y = `(sec^-1 "x")^2 ,"x" > 0`
⇒ `(d"y")/(d"x") = 2 sec^-1 "x"· (d(sec^-1"x"))/(d"x")`
⇒ `(d"y")/(d"x") = 2 sec^-1 "x"·(1)/(xsqrt(x^2 - 1))` ......(i)
⇒ `(d^2y)/(dx^2) = 2[1/(x^2(x^2 - 1))] + 2sec^-1x[[-sqrt(x^2 - 1) - x((2x)/(2sqrt(x^2 - 1))))/(x^2(x^2 - 1))]`
⇒ `(d^2"y")/(d"x"^2) = 2 [(1)/("x"^2("x"^2 -1)]] + 2 sec^-1 "x"· (1)/(xsqrt("x"^2 - 1)) [ ("x"(1 - 2"x"^2))/("x"^2 ("x"^2 - 1))] ` .......(ii)
From (i) and (ii), we get
`(d^2"y")/(d"x"^2) = 2 [(1)/("x"^2("x"^2 -1)]] + (d"y")/(d"x") [ ("x"(1 - 2"x"^2))/("x"^2 ("x"^2 - 1))] `
⇒ `"x"^2 ("x"^2 -1) (d^2"y")/(d"x"^2) + (2"x"^3 - "x")· (d"y")/(d"x") - 2 = 0`
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