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Question

An integrating factor of the differential equation xdydx+ylogx=xexx12,(x>0), is

A
xlogx
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B
(x)logx
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C
(e)(logx)2
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D
ex2
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Solution

The correct option is C (e)(logx)2
Given differential equation is
xdydx+ylogx=xexx(1/2)logx
or dydx+y1xlogx=exx(1/2)logx
Here, P=1xlogx and Q=exx(1/2)logx
IF=e1xlogxdx
Put logx=t
1xdx=dt
IF=etdt
IF=et22
=e(logx)22=(e)(logx)2.

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