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Question
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 + eu| = log c, where c1 = log c
∴ log |y (u + eu)| = log c
∴ y(u + eu) = c
∴ `"y"("x"/"y" + "e"^("x"/"y")) = "c"`
∴ x + `"ye"^("x"/"y") = c`
This is the general solution.
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