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Question
Solve the following homogeneous differential equation:
(y2 – 2xy) dx = (x2 – 2xy) dy
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Solution
`((y^2 - 2xy))/((x^2 - 2xy)) = ("d"y)/("d"x)`
⇒ `("d"y)/("d"x) = ((y^2 - 2xy))/((x^2 - 2xy))` ........(1)
It is a homogeneous differential equation, same degree in x and y.
Put y = vx and `("d"y)/("d"x) = "v" + x "dv"/("d"x)`
Equation (1)
⇒ `"v" + x "dv"/("d"x) = ([("v"x)^2 - 2x("v"x)])/([x^2 - 2x("v"x)])`
`"v" + x "dv"/("d"x) = ("v"^2x^2 - 2"v"x^2)/([x^2 - 2"v"x^2])`
`"v" + x "dv"/("d"x) = (x^2("v"^2 - 2"v"))/(x^2(1 - 2"v"))`
`"v" + x "dv"/("d"x) = (("v"^2 - 2"v"))/((1 - 2"v"))`
`x "dv"/("d"x) = (("v"^2 - 2"v"))/((1 - 2"v")) - "v"`
`x "dv"/("d"x) = (("v"^2 - 2"v") - "v"(1 - 2"v"))/((1 - 2"v"))`
= `("v"^2 - 2"v" - "v" + 2"v"^2)/((1 - 2"v"))`
`x "dv"/("d"x) = ((3"v"^2 - 3"v"))/((1 - 2"v")) = (3"v"("v" - 1))/((1 - 2"v"))`
⇒ `x "dv"/("d"x) = (3"v"("v" - 1))/(-(2"v" - 1))`
`((2"v" - 1))/(3"v"("v" - 1)) "dv" = (-1)/x "d"x`
Integrating on both sides
`int (2"v" - 1 - 1 + 1)/(3"v"("v" - 1)) "dv" = - int 1/x "d"x`
`int [((2"v" - 2))/(3"v"("v" - 1)) + 1/(3"v"("v" - 1))] "dv" = - logx + log"c"`
`2/3 int (("v" - 1))/("v"("v" - 1)) "dv" + 1/3 int 1/("v"("v" - 1)) "dv" = log("c"/x)`
`2/3 int 1/"v" "dv" + 1/3 int (1/(("v" - 1)) - 1/"v") "dv" = log("c"/x)`
`2/3 log "v" + 1/3 (log("v" - 1) - log"v") = log("c"/x)`
`1/3 [2log "v" + log((("v" - 1))/"v")] = log("c"/x)`
`1/3 [log("v"^2 xx (("v" - 1))/"v")] = log("c"/x)`
`1/3 [log("v"("v" - 1))] = log("c"/x)`
`log("v"^2 - "v")^(1/3) = log ("c"/x)`
⇒ `("v"^2 - "v")^(1/3) = ("c"/x)`
By Partial Fraction
`1/("v"("v" - 1)) = "A"/(("v" - 1)) + "b"/(("v" - 1))`
`1/("v"("v" - 1)) = ("Av" "b"("v" - 1))/("v"("v" - 1))`
1 = Av + B(v – 1)
Put v = 1
A = 1
Put v = 0
B = – 1
∴ `1/("v"("v" - 1)) = 1/(("v" - 1)) - 1/"v"`
Cubing on both sides
`[("v"^2 - "v")^(1/3)] = ("c"/x)^3`
`("v"^2 - "v") = "c"^3/x^3`
`[(y/x)^2 - (y/x)] = "c"/x^3`
`(y^2/x^2 - y/x) = "c"/x^3`
`((y^2 - xy)/x^2) = "c"/x^3`
`(x^3(y^2 - xy))/x^2` = c
⇒ `x(y^2 - xy)` = c
⇒ `- x(xy - y^2)` – c ......[Take – c = c]
⇒ `x(xy - y^2)` = c
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