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Question

Solve:
dydx=x+y+1x+y1 when y=13 at x=23

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Solution

Given,

dydx=x+y+1x+y1

substitute v=x+ydvdx=1+dydx

dvdx1=v+1v1

dvdx=v+1v1+1

dvdx=2vv1

v12vdv=dx

integrating on both sides, we get,

12(vlogv)=x+c

12(x+ylog(x+y))=x+c

given, y=13,x=23

12(23+13log(23+13))=23+c

12(1log1)=23+c

1223=cc=16

12(x+ylog(x+y))=x16

6x+6y6log(x+y)=2x2

4x+6y6log(x+y)=2

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