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Q.

If x= sint and y= sinpt then the value of 1−x2dydx2−xdydx+p2y=

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a

0

b

1

c

2

d

-1

answer is A.

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Detailed Solution

dxdt=cos⁡tdydt=pcos⁡pt⇒dydx=pcos⁡ptcos⁡t=p1−sin2⁡pt1−sin2⁡tdydx=p1−y21−x2dydx21−x2=p21−y2 Diff 2dydxd2ydx21−x2+dydx2(−2x)=p2(−2y)dydxd2ydx21−x2−xdydx=−p2y1−x2d2ydx2−xdydx+p2y=0
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