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Question

The solution of the differential equation dydx=x+yx satisfying the condition y(1)=1 is:

A
y=xe(x1)
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B
y=xlnx+x
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C
y=lnx+x
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D
y=xlnx+x2
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Solution

The correct option is B y=xlnx+x
dydx=x+yx
Put y=vxdydx=v+xdvdx
dv=dxxv=logx+cyx=logx+c
Since y(1)=1
y=xlogx+x

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