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Question

Find the general solution of the differential equation xdydx+2y=x2logx.

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Solution

We have,

xdydx+2y=x2logx

dydx+2yx=x2logxx

dydx+2yx=xlogx

Comparing this equation

dydx+Py=Q

Then, P=2x,Q=xlogx

I.F.=ePdx

=e2xdx

=e2logx

=elogx2elogx=x

=x2

Then, Complete solution is

yI.F.=I.F.Qdx+C

yx2=x2.xlogxdx+C

yx2=x3logxdx+C

Now,

x3logxdx=logxx3dx(dlogxdxx3dx)dx+C

=x4logx4x2dx+C

=x4logx4x33+C

yx2 = x4logx4x33+C

y= xlogx4x3+C

Hence, this is the answer.

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