Question

Solve the following differential equation.

x²y dx − (x³ + y³) dy = 0

Answer 1

Given: x²y dx − (x³ + y³) dy = 0

∴ x²y dx = (x³ + y³) dy

∴ `dy/dx = (x^2y)/(x^3+y^3)` ...(i)

Which is a homogeneous differential equation.

Put y = tx  ...(ii)

Differentiating w.r.t. x, we get

`dy/dx = t + x dt/dx` ...(iii)

Substituting (ii) and (iii) in (i), we get

`t + x dt/dx = (x^2 * tx)/(x^3 + t^3 x^3)`

∴ `t + x dt/dx  = (x^3 * t)/(x^3(1 + t^3))`

∴ `x dt/dx =  t /(1 + t ^3) - t`

∴ `x dt/dx = (t - t - t^4)/(1 + t^3)`

∴ `x dt/dx = (-t^4)/(1 + t^3)`

∴ `(1 + t^3)/t^4  dt = -dx/x`

Which is in variable separable form.

Integrating on both sides, we get

∴ `int(1 + t^3)/t^4dt = - int 1/x dx`

∴ `int(1/t^4 + 1/t)dt = - int1/x dx`

∴ `int t^-4 dt + int 1/t dt = - int 1/x dx`

∴ ` t^-3/-3 + log |t| = - log |x| + c`

∴ `- 1/(3t^3) + log t = - log x + c`

Resubstituting the value of `t = y/x`, we get

∴ `-1/3 * 1/(y/x)^3 + log (y/x) = - log x + c`

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

∴ `log y - x^3/(3y^3) = c` 

∴ `log y - x^3/(3y^3) = c` is the required general solution.