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Question

Find a particular solution of the differential equation (xy)(dx+dy)=dxdy given that y = -1, when x = 0.

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Solution

Given, differential equation is (xy)(dx+dy)=dxdy
dx+dy=dxdyxy
On integrating both sides, we get (dx+dy)=dxdyxy+C
Let xy=tdxdy=dt
dx+dyC=dxdyxy=dtt=log t=log|xy|x+y=log|xy|+C(i)
It is given that when x = 0, y = -1
0+(1)=log(0+1)+CC=1
On substituting this value in Eq. (i), we get the required particular solution as x+y=log|xy|1log|xy|=x+y+1


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