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Question
Solve the following differential equation:
`"dy"/"dx" + ("x" - "2y")/("2x" - "y") = 0`
Solution
`"dy"/"dx" + ("x" - "2y")/("2x" - "y") = 0`
∴ `"dy"/"dx" = - (("x" - "2y")/("2x" - "y"))` ....(1)
Put y = vx
∴ `"dy"/"dx" = "v + x""du"/"dx"`
∴ (1) becomes, `"v + x" "du"/"dx" = - (("x" - "2vx")/("2x" - "vx"))`
∴ `"v + x""du"/"dx" = - ((1 - "2v")/(2 - "v"))`
∴ `"x" "dv"/"dx" = - ((1 - "2v")/(2 - "v")) - "v"`
∴ `"x" "dv"/"dx" = (- 1 + "2v" - 2"v" + "v"^2)/(2 - "v")`
∴ `"x" "dv"/"dx" = ("v"^2 - 1)/(2 - "v")`
∴ `(2 - "v")/("v"^2 - 1)"dv" = 1/"x" "dx"`
Integrating both sides, we get
`int (2 - "v")/("v"^2 - 1)"dv" = int 1/"x" "dx"`
∴ `2 int 1/("v"^2 - 1) "dv" - 1/2 int "2v"/("v"^2 - 1)"dv" = int 1/"x" "dx"`
∴ `2 xx 1/2 log |("v" - 1)/("v" + 1)| - 1/2 log |"v"^2 - 1| = log |"x"| + log "c"_1` .....`[because "d"/"dx" ("v"^2 - 1) = "2v" and int("f"'("x"))/("f"("x")) "dx" = log |"f"("x")| + "c"]`
∴ `log |("v" - 1)/("v" + 1)| - log |("v"^2 - 1)^(1/2)| = log |"c"_1 "x"|`
∴ `log |("v" - 1)/("v" + 1) . 1/sqrt("v"^2 - 1)| = log |"c"_1 "x"|`
∴ `("v" - 1)/("v" + 1) . 1/sqrt("v"^2 - 1) = "c"_1 "x"`
∴ `("y"/"x" - 1)/("y"/"x" + 1) . 1/(sqrt("y"^2/"x"^2 - 1)) = "c"_1 "x"`
∴ `("y" - "x")/("y" + "x") . "x"/sqrt("y"^2 - "x"^2) = "c"_1"x"`
∴ `("y" - "x")/("y" + "x") = "c"_1 sqrt("y"^2 - "x"^2)`
∴ `("y" - "x")/("y" + "x") = "c"_1 sqrt(("y" - "x")("y" + "x")`
∴ `sqrt("y" - "x") = "c"_1 ("y" + "x")^(3/2)`
∴ y – x = `"c"_1^2 ("x + y")^3`
∴ y – x = c(x + y)3, where c = `"c"_1^2`
∴ y = c(x + y)3 + x
This is the general solution.
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