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Question

The solution of the differential equation xdydx=y1+logx is :

A
y=logx+C
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B
y=C1+logx
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C
y=C(x+logx)
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D
y=x+log(Cx)
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E
y=C(1+logx)
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Solution

The correct option is D y=C(1+logx)

xdydx=y1+logxdyy=dxx(1+logx)

Put t=logx

dtdx=1xdt=dxxdyy=dt1+tdyy=dt1+tlogy=log(1+t)+logClogy=logC(1+t)y=C(1+t)

Substituting t

y=C(1+logx)

So option E is correct.


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