Solve `(1+e^((x)/(y)))dx + e^((x)/(y)) (1-(x)/(y))dy = 0`

Explanation

` (1+e^(x//y))dx + e^(x//y) (1-x/y) dy =0`
` (1+e^(x//y))(dx)/(dy) +e^(x//y) (1-x/y) =0`
Now ` x/y =u " " therefore x = uy " " therefore (dx)/(dy) = u+ y (du)/(dy)`
` thereforer ` (1) becomes, ` (1+e^(u)) (u+y(du)/(dy)) +r^(u) (1-u) =0`
` u + ue^(u) +y(1+e^(u) (du)/(dx) +e^(u) -ue^(u) =0`
` (u +e^(u)) + y(1+e^(u)) (du)/(dy) =0 " " therefore (dy)/y + (1 +e^(u))/(u + e^(u)) du =0`
` int (dy)/(y) + int (1+e^(u))/(u+e^(u)) du =c_(1)`
` d/(du) (u+e^(u)) =1 +e^(u)`
from (2)/ the general solution is
` log |y| + |log | u +e^(u) = log c , " where " c_(1) = log c`
` log |y( u +e^(u))| = log c therefroe x+ye^(x//y) =c`
This is the general solution.