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Question

If y=f(x)=(x+1+x2)n, then (1+x2)d2ydx2+xdydx equals

A
f(x)
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B
0
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C
n2f(x)
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D
nf(x)
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Solution

The correct option is C n2f(x)
Given equation: y=f(x)=(x+1+x2)n

y=n(x+1+x2)n1(1+2x21+x2)

=n(x+1+x2)n1((x+1+x2)1+x2)

=n(x+1+x2)n1+x2

=ny1+x2(1)

y′′=ny1+x2nyx(1+x2)32(2)

now putting these values in the equation, we get
=(1+x2)y′′+xy=(1+x2)⎜ ⎜ ⎜ ⎜ny1+x2nyx(1+x2)32⎟ ⎟ ⎟ ⎟+x(ny1+x2)

=(1+x2)nynyx1+x2+nyx1+x2

=(1+x2)ny=n2y=n2f(x)

So, option (c) is correct.

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