**Solve the following differential equation:**

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

#### Solution

`(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")"dy" = 0` .....(1)

Put `"x"/"y" = "u"`

∴ x = uy

∴ `"dx"/"dy" = "u + y""du"/"dy"`

∴ (1) becomes, `(1 + "e"^"u")("u + y""du"/"dy") + "e"^"u"(1 - "u") = 0`

∴ `"u" + "ue"^"u" + "y"(1 + "e"^"u")"du"/"dy" + "e"^"u" - "ue"^"u" = 0`

∴`("u" + "e"^"u") + "y"(1 + "e"^"u")"du"/"dy" = 0`

∴ `"dy"/"y" + (1 + "e"^"u")/("u" + "e"^"u")"du" = 0`

∴ `int "dy"/"y" + int(1 + "e"^"u")/("u" + "e"^"u")"du" = "c"_1` ....(2)

∴ `"d"/"du" ("u" + "e"^"u") = 1 + "e"^"u" and int("f"'("u"))/("f"("u")) "du" = log |"f"("u")| + "c"`

∴ 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

∴ y(u + e^{u}) = c

∴ `"y"("x"/"y" + "e"^("x"/"y")) = "c"`

∴ x + `"ye"^("x"/"y") = c`

This is the general solution.