Solve the differential equation `y e^(x/y)dx=(x e^(x/y)+y^2)dy(y!=0)`

Explanation

Given differential equation is
`ye((x)/(y))dx=(xe^((x)/(y))+y^(2))dy`
`implies e^((x)/(y))(dx)/(dy)=(x)/(y)e((x)/(y))+yimpliese^((x)/(y))(dx)/(dy)-(x)/(y)e^((x)/(y))=y` ………`(1)`
Let `(x)/(y)=vimpliesx=vyimplies(dx)/(dy)=v+y(dv)/(dx)`
From equation `(1)`, `e^(v)(v+y(dv)/(dy))-ve^(v)=y`
`implies e^(v)y(dv)/(dy)=yimpliese^(v)(dv)/(dy)=1impliese^(v)dv=dy`
On integration,
`inte^(v)dv=int1dyimpliese^(v)=y+Cimpliese^((x)/(y))=y+C`