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

2yexydx+(y-2xexy)dy=0
⇒2exy=-(1-2xyexy)dy/dx
Put xy=t
x=y/t
1=dy/dx t+y dt/dy
⇒dy/dx=1/t(1-y dt/dx)
⇒2et=-(1-2tet)1/t(1-y dt/dx)
⇒2tet=(2tet-1)(1-y dt/dx)=2tet-1-y2tetdt/dx+y dt/dx
⇒1=y(1-2tet)dt/dx=x/t(1-2tet)dt/dx
∫dx/x=∫(1/t-2et)dt
⇒lnx=lnt-2et+C
⇒lnx=lnxy-2exy+C
⇒lny=-2exy+C
At x=0,y=1
⇒0=-2e0+C
⇒C=2
lny=2-2exy