Question

Solve the following differential equation

x²y dx – (x³ + y³)dy = 0

Answer 1

x²y dx – (x³ + y³)dy = 0

⇒ (x³ + y³).dy = x²y.dx

⇒ `(dy)/(dx) = (x^2y)/(x^3 + y^3)`  ......(1)

Let y = vx  ......(2)

∴ `(dy)/(dx) = "v" + x (d"v")/(dx)` 

∴ Equation (1) can be written as

`"v" + x. (d"v")/(dx) = (x^2."v"x)/(x^3 + "v"^3x^3)`

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

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

⇒ `x. (d"v")/(dx) = "v"/(1 + "v"^3) - "v"`

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

⇒ `x. (d"v")/(dx) = - "v"^4/(1 + "v"^3)`

⇒ `(1 + "v"^3)/"v"^4 . d"v" = - 1/x. dx`

Integrating both sides. we get

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

⇒ `int (1/"v"^4 + 1/"v"). d"v" = - int 1/x. dx`

⇒ `int "v"^-4. d"v" + int 1/"v". d"v" = - int 1/x. dx`

⇒ `"v"^(-3)/(-3) + log |"v"| = - log |x| + C`

⇒ `- 1/3 . 1/"v"^3 + log |"v"| = - log |x| + C`

Resubstituting the value of v from (2),

⇒ `- 1/3 1/(y/x)^3 + log |y/x| = - log |x| + C`

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

∴ `log y = x^3/(3y^3) + C`