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Question

Solve: (x2+1)dydx+4xy=1x2+1

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Solution

Given,

(x2+1)dydx+4xy=1x2+1

dydx+4x(x2+1)y=1(x2+1)2

dydx+Py=Q

P=4x(x2+1),Q=1(x2+1)2

I.F=ePdx=e4x(x2+1)dx=(x2+1)2

y×I.F=Q×I.Fdx

y×(1+x2)2=1(x2+1)2×(1+x2)2dx

y(1+x2)2=dx

y(1+x2)2=x+c

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