Question

Solve the following differential equation:

`(y^2 - 2xy) "d"x = (x^2 - 2xy) "d"y`

Answer 1

Given equation is (y² – 2xy) dx = `(x^2 - 2xy) "d"y`

y² – 2xy = `(x^2 - 2xy) ("d"y)/("dx)`

∴ The equation can written as

`(y^2 - 2xy)/(x^2 - 2xy) = ("d"y)/("d"x)`

∴ `("d"y)/("d"x) = (y^2 - 2xy)/(x^2 - 2xy)`  .......(1)

This is a homogeneous differential equation

∴ Put y = vx

`("d"y)/("d"x) = "v"(1) + x "dv"/("d"x)`

∴ Equation (1) becomes,

`"v" + x "dv"/("d"x) = ("v"^2x^2 - 2x  "v"x)/(x^2 - 2x "v"x)`

= `("v"^2x^2 - 2x^2"v")/(x^2 - 2x^2"v")`

∵ y = vx

y = v²x²

`"v" + x "dv"/("d"x) = (x^2("v"^2 - 2"v"))/(x^2(1 - 2"v"))`

`x "dv"/("d"x) = ("v"^2 - 2"v")/(1 - 2"v") - "v"`

= `("v"^2 - 2"v" - "v"(1 - 2"v"))/(1 - 2"v")`

= `("v"^2 - 2"v" - "v" + 2"v"^2)/(1 - 2"v")`

`x"v"/("d"x) = (3"v"^3 - 3"v")/(1 - 2"v")`

`((1 - 2"v"))/(3"v"^2 - 3"v") "dv" = ("d"x)/x`

Multpily by – 3 on both sides, we get

`((-3 + 6"v"))/(3"v"^2 - 3"v") "dv" = - 3 ("d"x)/x`

`((6"v" - 3))/(3"v"^2 - 3"v") "dv" = - 3 ("d"x)/x`

Integrating on both sides, we get

`int (6"v" - 3)/(3"v"^2 - "v") "dv" = - 3 int ("d"x)/x`

log (3v² – 3v) = – 3 log x + log C

log (3v² – 3v) = – log x³ + log C

= log c – log x³ 

log (3v² – 3v) = `log "C"/x^3`

3v² – 3v = `"C"/x^3`

`3(y/x)^2 - 3(y/x) = "C"/x^3`

y = vx

v = `y/x`

`3 y^2/x^2 - 3y/x = "C"/x^3`

`(3y^2 - 3xy)/x^2 = "C"/x^3`

`(x^3(3y^2 - 3xy))/x^2 = "C"/x^3`

3xy² – 3x²y = C

3(xy² – x²y) = C

xy² – x²y = `"C"/3`

xy² – x²y = C

∵ C = `"C"/3`