Example:  Solve the following differential equation.

y⁽³⁾  -  12 y''+48 y' + 64 y = 12 - 32 e^-8t + 2 e^4t

Solution:   

We first require the complementary solution thus the characteristic equation is,

r³ - 12 r² + 48 r - 64 = (r - 4)3 = 0

r = 4; when multiply with 3

We've found a single root of multiplicity 3 thus the complementary solution is,

yc(t) = c₁ e^4t + c₂ t e^4t + c₃ t² e^4t

Here is, our first guess for an exact solution is,

Y_p = A + B e^-8t + C e^4t

Remember that the last term in our guess is into the complementary solution thus we'll require to add one at least one t to the third term in our guess. Also see that multiplying the third term with either t or t² will result into a new term which is until now in the complementary solution and therefore we'll require multiplying the third term by t³ in order to find a term that is not included in the complementary solution.

Our last guess is,

Y_p = A + B e^-8t + C t³ e^4t

Here all we require to do is take three derivatives of this, plug that in the differential equation and simplify to find (we'll leave this to you to verify the work now),

-64 A - 1728 B e^-8t + 6 C e^4t = 12 - 32 e ^-8t + 2 e^4t

By setting coefficients equal and solving provides,

t⁰:         -64 A = 12        ⇒         A = -(3/16)

e^-8t:      -1728 B = -32 ⇒         B = 1/54

e^4t:       6C = 2              ⇒         C = 1/3

An exact solution is then,

Y_p = -(3/16) + (1/54) e^-8t + (1/3) t³ e^4t

Then the general solution to above differential equation is,

y(t) = c₁ e^4t + c₂ t e^4t + c₃ t² e^4t -(3/16) + (1/54) e^-8t + (1/3) t³ e^4t

Okay, we've only worked one example here, but remember that we mentioned earlier that with the exception of the extension to the method that we used in this example the work here is identical to work we did the 2nd order material.