Question

Solve the following differential equation:

`(x^3 + y^3)"d"y - x^2 y"d"x` = 0

Answer 1

The given equation can be written as

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

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

This is a homogeneous equation

y = vx

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

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

= `(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"`

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

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

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

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

On integrating,

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

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

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

`int "v"^-4 "dv" + log "v" = - logx + log "c"`

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

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

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

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

`log "c" + 1/(3(y/x)^3) = log[y/x x]`

∵  y = vx

v = `y/x`

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

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

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

`log(y/"c") = x^3/(3y^3)`

`y/"c" = "e"^(x^3/(3y^3))`

y = `"ce"^(x^3/(3y^3))`