Question

Find the general solution of the differential equation: (x³ + y³)dy = x²ydx

Answer 1

Given differential equation is (x³ + y³)dy = x²ydx

∴ `(dx)/(dy) = (x^3 + y^3)/(x^2y)`  ...(i)

Put x = vy

⇒ `(dx)/(dy) = v + y (dv)/(dy)`

From equation, we have

`v + y (dv)/(dy) = ((vy)^3 + y^3)/((vy)^2y)`

`v + y (dv)/(dy) = (v^3y^3 + y^3)/(v^2y^3)`

`v + y (dv)/(dy) = (v^3 + 1)/v^2`

`y (dv)/(dy) = (v^3 + 1)/v^2 - v`

`y (dv)/(dy) = 1/v^2`

`v^2dv = (dy)/y`   ...(Variable seponation method)

Integrating both sides, we get

`int v^2dv = int (dy)/y`

`v^3/3` = log + C

Putting v = `x/y`, we get

`x^3/(3y^3)` = log + C