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Question

The solution of the differential equation xdydx=y(logylogx+1) is :

A
logy=cx +logx
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B
xlogy=cx
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C
xlogy=cy
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D
y=xelogx
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Solution

The correct option is B logy=cx +logx
xdydx=y(logylogx+1)
y=vx
dydx=v+xdvdx

v+xdvdx=yx(logyx+1)

v+xdvdx=v(logv+1)

v+xdvdx=vlogv+v

xdvdx=vlogv

dvvlogv=dxx
Integrating both sides
dvvlogv=dxx
log(v)=logx+logc
log(logyx)=logcx
logy=cx+logx
Hence, the answer is logy=cx+logx.


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