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Question
Solve the following differential equation:
`"dy"/"dx" = ("2y" - "x")/("2y + x")`
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Solution
`"dy"/"dx" = ("2y" - "x")/("2y + x")` ....(1)
Put y = vx ∴ `"dy"/"dx" = "v + x""dv"/"dx"`
∴ (1) becomes, `"v + x""dv"/"dx" = ("2vx - x")/("2vx + x")`
∴ `"v + x""dv"/"dx" = ("2v" - 1)/("2v" + 1)`
∴ `"x""dv"/"dx" = ("2v" - 1)/("2v" + 1) - "v" = ("2v" - 1 - "2v"^2 - "v")/("2v + 1")`
∴ `"x""dv"/"dx" = - (("2v"^2 - "v" + 1)/("2v" + 1))`
∴ `("2v" + 1)/("2v"^2 - "v" + 1) "dv" = - 1/"x" "dx"`
Integrating both sides, we get
`int ("2v" + 1)/("2v"^2 - "v" + 1) "dv" = - int 1/"x" "dx"`
∴ `int (1/2 ("4v" - 1) + 3/2)/("2v"^2 - "v" + 1) "dv" = - int 1/"x" "dx"`
∴ `1/2 int ("4v" - 1)/("2v"^2 - "v" + 1) "dv" + 3/2 int 1/("2v"^2 - "v" + 1) "dv" = - int 1/"x"`
∴ `1/2 int ("4v" - 1)/("2v"^2 - "v" + 1) "dv" + 3/4 int 1/("v"^2 - 1/2"v" + 1/2)"dv" = - int 1/"x" "dx"`
∴ `1/2 int ("4v" - 1)/("2v"^2 - "v" + 1) "dv" + 3/4 int 1/(("v"^2 - 1/2"v" + 1/16) + 7/16) "dv" = - int 1/"x" "dx"`
∴ `1/2 int ("4v" - 1)/("2v"^2 - "v" + 1) "dv" + 3/4int 1/(("v" - 1/4)^2 + (sqrt7/4)^2)"dv" = - int 1/"x" "dx"`
∴ `1/2 log |2"v"^2 - "v" + 1| + 3/4 xx 1/((sqrt7/4)) tan^-1 |("v" - 1/4)/((sqrt7/4))| = - log |x| + "c"_1 .....[because "d"/"dv" (2"v"^2 - "v" + 1) = 4"v" - 1 and int ("f"'("v"))/("f"("v")) "dv" = log |"f"("v")| + c]`
∴ `1/2 log |2 ("y"^2/"x"^2) - "y"/"x" + 1| + 3/sqrt7 tan^-1 ((4("y"/"x") - 1)/sqrt7) = - log |"x"| + "c"_1`
∴ `1/2 log |(2"y"^2 - "xy" + "x"^2)/"x"^2| + 3/sqrt7 tan^-1 ((4"y - x")/(sqrt7"x")) = - log |"x"| + "c"_1`
∴ `log |("x"^2 - "xy" + "2y"^2)/"x"^2| + 6/sqrt7 tan^-1 (("4y - x")/(sqrt7"x")) = - 2 log |"x"| + 2"c"_1`
∴ `log |"x"^2 - "xy" + "2y"^2| - log"x"^2 + 6/sqrt7 tan^-1 (("4y - x")/(sqrt7"x")) = - log "x"^2 + "c"_1 "where" "c" = 2"c"_1`
∴ `log |"x"^2 - "xy" + "2y"^2| + 6/sqrt7 tan^-1 (("4y - x")/(sqrt7"x")) = "c"`
This is the general solution.
Notes
The answer in the textbook is incorrect.
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