Show that the differential equation `2y e^(x/y) dx+(y-2x e^(x y)`) `dy=0`ishomogeneous. Find the particular solution of this differential equation,given that `x=0`when `y=1.`

Show that the differential equation `2y e^(x/y) dx+(y-2x e^(x y)`) `dy

1.	Show that the differential equation `2y e^(x/y) dx+(y-2x e^(x y)`) `dy=0`ishomogeneous. Find the particular solution of this differential equation,given that `x=0`when `y=1.`
Answer» The given differential equation may be written as `(dx)/(dy)=(2xe^(x/y)-y)/(2ye^(x//y))`……………………(i) On dividing the Nr and Dr of RHS of (i) by y, we get `(dx)/(dy)={(2x/y.e^(x-y)-1)/(2e^(x//y))}=f(x/y)`…………….(ii) So, the given differential equation is homogeneous. Putting `x=vy` and `(dx)/(dy)=v+y(dv)/(dy)` in (ii), we get `v+y(dv)/(dy)=(2ve^(v)-1)/(2e^(v))` `rArr y(dv)/(dx)={(2ve^(v)-1)/(2e^(v))-v}=-1/(2e^(v))` `rArr 2e^(v)dv=-1/ydy` `rArr 2inte^(v)dy=-int1/ydy` [on integrating both sides] `rArr 2e^(v)=-log\|y\|+C` `rArr 2e^(x//y)+log\|y\|=C`...............(iii) `[therefore v=x/y]`. Putting, `y=1` and `x=0` in (iii), we get `C=2`. `therefore 2e^(x//y)+log\|y\|=2` is the required solution.