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Question

Solve the differential equationxdydx=y(logylogx+1)

A
logylogx=x+c
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B
logy+logx=cx.
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C
logylogx=cx.
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D
logylogx+cx=0.
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Solution

The correct option is C logylogx=cx.
xdydx=y(logylogx+1)
1ydydx1xlogy=1logxx.
Put logy=v1ydy=dv
dvdxvx=1logxx ...(1)
Here P=1xPdx=1xdx=logx
I.F=elogx=elogx1=1x
Multiplying (1) by I.F, we get
1xdvdxvx2=1logxx2
Integrating both sides we get
1xv=logxx+c
logylogx=cx

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