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Question

Find the particular solution of the differential equation
2yexydx+(y2xexy)dy=0 given that x=0 when y=1.

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Solution

2yexydx+(y2xexy)dy=0
2exy=(12xyexy)dydx
Put xy=t
x=yt
1=dydxt+ydtdy
dydx=1t(1ydtdx)
2tet=(12tet)1t(1ydtdx)
2tet=(2tet1)(1ydtdx)=2tet1y2tetdtdx+ydtdx
1=y(12tet)dtdx=xt(12tet)dtdx
dxx=(1t2et)dt
lnx=lnt2et+C
lnx=lnxy2exy+C
lny=2exy+C
At x=0,y=1
0=2e0+CC=2
lny=22exy

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