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Question

Solve the differential equation (x+y)dydx=1.

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Solution

We need to solve (x+y)dydx=1

Consider (x+y)dydx=1

dydx=1x+y

dxdy=x+y

dxdyx=y

Comparing with dxdy+px=q we get p=1,q=y

Integrating factor I.F=epdy=e1dy=ey

General solution is given by xepdy=qepdy+c

Hence the solution is xey=yeydy+c...(1)

Consider yeydy

Let u=y,v=ey
du=dy,dv=eydy

yeydy=yey+eydy

yeydy=yeyey

Thus (1) becomes,

xey=yeyey+c

Divide throughout by ey we get

x=y1+cey

x+y+1=cey

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