Question

Solve the following homogeneous differential equation:

(y² – 2xy) dx = (x² – 2xy) dy

Answer 1

`((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