Question

Solve the following homogeneous differential equation:

`("d"y)/("d"x) = (3x - 2y)/(2x - 3y)`

Answer 1

`("d"y)/("d"x) = (3x - 2y)/(2x - 3y)`  ......(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) = (3x - 2("vx"))/(2x - 3("vx"))`

`"v" + x "dv"/("d"x) = (x[3 - 2"v"])/(2[2 - 3"v"])`

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

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

= `(3 - 2"v" - 2"v" + "v"^2)/((2 - "v"))`

`x "dv"/("d"x) = ((3"v"^2 - 4"v" + 3))/(-(3"v" - 2))`

`((3"v" - 2))/((3"v"^2 - 4"v" + 3)) "dv" = - 1/x  "d"x`

Integrating on both sides

`int ((3"v" - 2))/((3"v"^2 - 4"v" + 3))  "dv" = - int 1/x  "d"x`

`1/2 int ((6"v" - 4))/((3"v"^2 - 4"v" + 3)) = - int 1/x  "d"x`

`1/2 log(3"v"^2 - 4"v" + 3) = - log x + log "c"`

`log (3"v"^2 - 4"v" + 3)^(1/2) = log("c"/x)`

⇒ `(3"v"^2 - 4"v" + 3)^(1/2) = "c"/x`

`sqrt(3(y/x)^2 - 4(y/x) + 3) = "c"/x`

`sqrt((3y^2)/x^2 - (4y)/x + 3) = "c"/x`

`sqrt((3y^2 - 4xy + 3x^2)/x) = "c"/x`

Squaring on both sides

`(3y^2 - 4xy + 3x^2)/x^2 = ("c"/x)^2`

`3y^2 - 4xy + 3x^2 = x^2 xx ("c"^2/x^2)`

`3y^2 - 4xy + 3x^2 = "c"^2`

⇒ `3y^2 - 4xy + 3x^2` = c