The ODE,
y''' + 3y'' - 4y' - 12y = 0
has characteristic equation
r ³ + 3r ² - 4r - 12 = 0
which can be factorized as
r ² (r + 3) - 4 (r + 3) = (r ² - 4) (r + 3) = 0
so that its roots are r = ±2 and r = -3.
Then the characteristic solution to the ODE is
y(x) = C₁ e ⁻²ˣ + C₂ e ²ˣ + C₃ e ⁻³ˣ
