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प्रश्न
Find the particular solution of the following differential equation:
`2e ^(x/y) dx + (y - 2xe^(x/y)) dy = 0," When" y (0) = 1`
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उत्तर
`(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 + 2ev| = - 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 + 2ev| + log y = log c
∴ log |y (v + 2ev)| = log c
∴ y(v + 2ev) = c
∴ `"y"("x"/"y" + 2"e"^("x"//"y"))`= c
∴ x + 2yex/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 + 2yex/y = 2
Notes
The question is modified.
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