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प्रश्न
If x = sin t, y = sin pt, prove that`(1-"x"^2)("d"^2"y")/"dx"^2 - "x" "dy"/"dx" + "p"^2"y" = 0`
उत्तर
x = sin t
`⇒ "dx"/"dt" = cos"t"`
y = sin pt
`⇒ "dy"/"dt" = "p cos pt"`
`"dy"/"dx" = ("dy"/"dt")/("dz"/"dt") = "p cos pt"/"cos t" = ("p"sqrt(1-sin^2"pt"))/(sqrt(1-sin^2"t"))`
`⇒ "dy"/"dx" = ("p"sqrt(1-"y"^2))/sqrt(1-"x"^2)`
Squaring both sides we get
`("dy"/"dx")^2 = ("p"^2(1-"y"^2))/((1-"x"^2))`
`⇒ (1-"x"^2)("dy"/"dx")^2 = "p"^2 (1-"y"^2)`
Differentiating with respect to x
`(1-"x"^2)"d"/"dx"("dy"/"dx")^2 + ("dy"/"dx")^2 "d"/"dx" (1-"x"^2) = "p"^2 (-2"y" "dy"/"dx")`
`⇒(1-"x"^2)2("dy"/"dx") (("d"^2"y")/"dx"^2) + ("dy"/"dx")^2 (-2"x") = -2"yp"^2 "dy"/"dx"`
`⇒ 2 ("dy"/"dx") {(1-"x"^2) (("d"^2"y")/"dx"^2) - "x" ("dy"/"dx") + "yp"^2} = 0`
Either `2 ("dy"/"dx") = 0` or `{(1-"x"^2)(("d"^2"y")/"dx"^2) - "x" ("dy"/"dx") + "yp"^2} = 0`
But ` 2 ("dy"/"dx") ≠ 0`
So, `{(1-"x"^2) (("d"^2"y")/"dx"^2) - "x" ("dy"/"dx") + "yp"^2} = 0`
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