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Question

If x=sin t,y=sin kt,show that (1x2)d2ydx2xdydx+k2y=0.

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Solution

Here x=sint t,y=sin kt
dxdt=cos t,dydt=k cos kt dydx=kcos ktcos t
cos tdydx=k cos kt cos2 t(dydx)2=k2 cos2 kt
(1sin2t)(dydx)2=k2(1sin2 kt)(1x2)(dydx)2=k2(1y2)

Differentiating w.r.t. x both the sides,

(1x2)×2dydx(d2ydx2)+(dydx)2(2x)=2k2ydydx
(1x2)d2ydx2xdydx+k2y=0

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