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प्रश्न
If `"y" = (sin^-1 "x")^2, "prove that" (1 - "x"^2) (d^2"y")/(d"x"^2) - "x" (d"y")/(d"x") - 2 = 0`.
उत्तर
Here,
`"y" = (sin^-1 "x")^2`
Now,
`"y"_1 = 2 sin^-1 "x" (1)/(sqrt(1 - "x"^2)`
⇒ `"y"_2 = (2)/(1 - "x"^2) + (2"x" sin^-1 "x")/((1 - "x"^2)^(3/2)`
⇒ `"y"_2 = (2)/(1 - "x"^2) + (2"x" sin^-1 "x")/((1 -"x"^2) sqrt(1 - "x"^2)`
⇒ `"y"_2 = (2)/(1 -"x"^2) + ("xy"_1)/((1 - "x"^2)`
⇒ `"y"_2 (1 - "x"^2) = 2 + "xy"_1`
⇒ `"y"_2 (1 - "x"^2) - "xy"_1 - 2 = 0`
Therefore, `(1 - "x"^2) (d^2"y")/(d"x"^2) - "x" (d"y")/(d"x") - 2 = 0`
Hence Proved.
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