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Question

Solve the differential equation, dydx2xy1+x2=x2+1.

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Solution

Using the equation-

dydx+Py=Q

The solution is given by -

y(I.F)=Q(I.F)dx

where integration factor=I.F=ePdx

Here,

dydx2x1+x2y=x2+1

P=2x1+x2 and Q=1+x2

Pdx=2x1+x2dx=dzzdz

where z=x2+1dz=2xdx

=logz=log1z=log11+x2


I.F=ePdx=elog11+x2=11+x2

Required solution is-

y×11+x2=(x2+1)×11+x2dx=dx=x


y=x(x2+1)+c

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