Question

Solve x²ydx – (x³ + y³) dy = 0

Answer 1

x²ydx – (x³ + y³) dy = 0

x²ydx = (x³ + y³) dy

`("d"y)/("d"x) = (x^2y)/((x^3 + y^3))`  ......(1)

This is a homogeneous differential equation, same degree in x and y

Let y = vx and `("d"y)/("d"x) = "v" + x "v"/("d"x)`

Equation (1) 

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

⇒ `"v" + x  "dv"/("d"x) = (x^3"v")/(x^3(1 + "v"^3))`

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

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

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

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

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

Integrating on both sides

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

`int "v"^-4 "dv" + int 1/"v" "dv" = - log x +"c"`

`[("v"^(-4 + 1))/(-4 + 1)] + log "v" + log x = "c"`

`["v"^-3/(-3)] + log "v"x + "c"`

`-1/(3"v"^3) + log (y/x) + "c"`

`x^3/(-3y^3) + log y` = c

⇒ logy = `x^3/(3y^3) + "c"`