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Question

If y2=P(x) is a polynomial of degree 3, then
2ddx(y3d2ydx2) is equal to

A
P(x)+P(x)
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B
P(x) P(x)
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C
P(x) P′′′(x)
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D
a constant
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Solution

The correct option is C P(x) P′′′(x)
From y2=P(x), we have 2yy1=P(x), i.e.,
2y1=P(x)/y
2y2=yP′′(x)P(x)y1y2
=yP′′(x)P(x).P(x)/2yy2
=2y2P′′(x)(P(x))22y3
2y2y3=12[2P(x)P′′(x)(P(x))2]
Now, differentiating w.r.t. x on both sides,
2ddx(y3d2ydx2)=12[2{P(x) P′′(x)+P(x) P′′′(x)}2P(x) P′′(x)]
=12.2P(x) P′′′(x)=P(x) P′′′(x)

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