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Question

If the solution of differential equation x lnxdydx+y=2lnx is y(lnx)=(lnx)n+C then n= ___

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Solution

Type of the differential equation is what you have to identify. Here if you divide the given equation by xlnx, it becomes dydx+1xlnx×y=2x which when compared to standard form of linear differential equation matches it. Here P(x)=1xlnx

Q(x)=2x
Solution will be y×I.F=Q×(I.F.)dx+C
I.F.=eP(x)dx
So I.F=lnx.
So general solution will be
y×lnx=2xlnx dx+C
ylnx=(lnx)2+C
So n=2.

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