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

Explanation

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.