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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 - Mathematics

Question

Sum

`"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`

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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