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Question
Solve the following differential equation:
`(1 + 2"e"^("x"/"y")) + 2"e"^("x"/"y")(1 - "x"/"y") "dy"/"dx" = 0`
Solution
`(1 + 2"e"^("x"/"y")) + 2"e"^("x"/"y")(1 - "x"/"y") "dy"/"dx" = 0`
∴ `(1 + 2"e"^("x"/"y")) + 2"e"^("x"/"y")(1 - "x"/"y") * 1/("dy"/"dx") = 0`
∴ `(1 + 2"e"^("x"/"y")) "dx"/"dy" + 2"e"^("x"/"y")(1 - "x"/"y") = 0` ....(1)
Put `"x"/"y" = "u"`
∴ x = uy
∴ `"dx"/"dy" = "u + y""du"/"dy"`
∴ (1) becomes, `(1 + 2"e"^"u")("u" + "y""du"/"dy") + 2"e"^"u" (1 - "u") = 0`
`"u" + 2"ue"^"u" + "y"(1 + 2"e"^"u") "du"/"dy" + 2"e"^"u" - 2"ue"^"u" = 0`
∴ `("u" + "2e"^"u") + "y"(1 + 2"e"^"u")"du"/"dy" = 0`
Integrating both sides, we get
`int1/"y" "dy" + int(1 + 2"e"^"u")/("u" + 2"e"^"u") "du" = "c"_1`
∴ log |y| + log |u + 2eu| = log c, where c1 = log c ......`[because "d"/"du"("u" + 2"e"^"u") = 1 + 2"e"^"u" and int("f"'("u"))/("f"("u")) "du" = log |"f"("u") + "c"|]`
∴ log |y (u + 2eu)| = log c
∴ y(u + 2eu) = c
∴ `"y"("x"/"y" + 2"e"^("x"/"y"))` = c
∴ x + `2"y""e"^("x"/"y") = "c"`
This is the general solution.
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