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