Question

Find the particular solution of the following differential equation:

`2e ^(x/y) dx + (y - 2xe^(x/y)) dy = 0," When" y (0) = 1`

Answer 1

`(1 + 2"e"^("x"//"y"))"dx" + 2"e"^("x"//"y")(1 - "x"/"y")"dy" = 0`

∴ `(1 + 2"e"^("x"//"y"))"dx" = - 2"e"^("x"//"y")(1 - "x"/"y")"dy"`

∴ `(1 + 2"e"^("x"//"y"))"dx" = 2"e"^("x"//"y")("x"/"y" - 1)"dy"`

∴ `"dy"/"dx" = (2"e"^("x"//"y")("x"/"y" - 1))/(1 + 2"e"^("x"//"y"))`   .....(1)

Put x = vy

∴ `"dx"/"dy" = "v" + "y" "dv"/"dy"`

∴ (1) becomes, `"v" + "y" "dv"/"dy" = (2"e"^"v"("v - 1"))/(1 + "2e"^"v")`

∴ `"y" "dv"/"dy" = (2"e"^"v"("v - 1"))/(1 + "2e"^"v") - "v"`

`= (2"ve"^"v" - 2"e"^"v" - "v" - 2"ve"^"v")/(1 + "2e"^"v")`

`= - (("v" + 2"e"^"v")/(1 + "2e"^"v"))`

∴ `((1 + 2"e"^"v")/("v" + 2"e"^"v"))"dv" ≡ - 1/"y" "dy"`

Integrating both sides, we get

`int ((1 + 2"e"^"v")/("v" + 2"e"^"v"))"dv" ≡ - int 1/"y" "dy"`

∴ log |v + 2e^v| = - log y + log c  ....`[because "d"/"dx" ("v" + "2e"^"v") = 1 + 2"e"^"v" and int("f"'("v"))/("f"("v")) "dv" = log |"f"("v")| + "c"]`

∴ log |v + 2e^v| + log y = log c

∴ log |y (v + 2e^v)| = log c

∴ y(v + 2e^v) = c

∴ `"y"("x"/"y" + 2"e"^("x"//"y"))`= c

∴ x + 2ye^x/y = c

This is the general solution.

Now, y(0) = 1, i.e. when x = 0, y = 1

∴ 0 + 2(1)e⁰ = c

∴ c = 2

∴ the particular solution is x + 2ye^x/y = 2