मराठी

If x = sint and y = sin pt, prove that dydxdydxp(1-x2)d2ydx2-xdydx+p2y = 0

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प्रश्न

If x = sint and y = sin pt, prove that `(1 - x^2) ("d"^2"y")/("dx"^2) - x "dy"/"dx" + "p"^2y` = 0

बेरीज
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उत्तर

Given that: x = sin t and y = sin pt

Differentiating both the parametric functions w.r.t. t

`"dx"/"dt"` = cos t and `"dy"/"dt"` = cos pt. p = p cos pt

`"dy"/"dx" = ("dy"/"dt")/("dx"/"dt")`

= `("p" cos "pt")/cos "t"`

∴ `"dy"/"dx" = ("p" cos "pt")/cos"t"`

Again differentiating w.r.t. x,

`"d"/"dx"("dy"/"dx") = "p"*"d"/"dx"(cos"pt"/cos"t")`

⇒ `("d"^2y)/("dx"^2) = "p"[(cos "t" * "d"/"dx" (cos "pt") - cos "pt" * "d"/"dx" (cos "t"))/(cos^2"t")]`

= `"p"[(cos"t"(- sin "pt") * "p" "dt"/"dx" - cos "pt"(- sin "t") * "dt"/"dx")/(cos^2"t")]`

= `"p"[(-"p" cos "t" sin "pt" + cos "pt" sin "t")/(cos^2"t")]"dt"/"dx"`

= `"p"[(-"p" cos "t" sin "pt" + cos "pt" sin "t")/(cos^2"t")]* 1/cos"t"`

= `"p"[(-"p" cos "t" sin "pt" + cos "pt" sin "t")/cos^3"t"]`

Now we have to prove that

`(1 - x^2) * ("d"^2y)/("dx"^2) - x * "dy"/"dx" + "p"^2y` = 0

L.H.S. = `(1 - x^2) ["p"((-"p" cos "t" sin "pt" + cos "pt" sin "t")/cos^3"t")] - x."p" (cos "pt")/cos"t" + "p"^2y`

⇒ `(1 - sin^2"t") ["p"((-"p" cos "t" sin "pt" + cos "pt" sin "t")/cos^3"t")] - ("p" sin "t" * cos "pt")/cos"t" + "p"^2 * sin "pt"`

⇒ `cos^2"t" [(-"p"^2 cos "t" sin "pt" + cos "pt" sin "t")/(cos^3"t")] - ("p" sin "t" * cos "pt")/cos"t" + "p"^2 * sin "pt"`

⇒ `(-"p"^2 cos "t" sin "pt" + "p" cos "pt" sin "t")/cos "t" - ("p" sin "t" cos "pt")/cos"t" + "p"^2 sin "pt"`

⇒ `(-"p"^2 cos "t" sin "pt" + "p" cos "pt" sin "t" - "p" sin "t" cos "pt" + "p"^2 sin "pt" cos "t")/cos "t"`

⇒ `0/cos"t"` = 0 = R.H.S.

Hence proved.

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पाठ 5: Continuity And Differentiability - Exercise [पृष्ठ ११३]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
पाठ 5 Continuity And Differentiability
Exercise | Q 81 | पृष्ठ ११३

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