Solve the following initial value problem

y⁽³⁾ - 5y'' - 22y' + 56 y= 0;       y(0) = 1;           y'(0) = -2;         y''(0) = -4

Solution:

r³ - 5r² - 22r + 56 = (r + 4)(r - 2)(r - 7) = 0

= r₁ = -4, r₂ = 4 and r₃ = 7

Therefore we have three real distinct roots now and thus the general solution is,

y(t) = c₁ e^-4t + c₂ ex^2t + c₃ ex^7t

Differentiating a couple of times and applying the initial conditions provides the subsequent system of equations that we'll require solving in order to get the coefficients.

1 = y(0) = c₁ + c₂ + c₃

-2 = y'(0) = - 4c₁ + 2 c₂ + 7 c₃

-4 = y''(0) = 16 c₁ + 4 c₂ + 49 c₃

After solving all these equations we get:

c₁ = 14/33

c₂ = 13/15

c₃ = -(16/55)

Then the actual solution is,

 y(t) = (14/33) e^-4t + (13/15) ex^2t -(16/55)ex^7t

Therefore, outside of requiring to solve a cubic polynomial that we left the details for you to verify and requiring to solve a system of 3 equations to get the coefficients that we also left for you to fill the details, the work currently is pretty much identical to the work we did in solving a 2^nd order initial value problem.

Since the initial condition work is identical to work which we should be extremely familiar with to that point with the exception which it involved solving larger systems we're going to not bother with solving initial value problems for the rest of the examples. The basic point of this section is the new concepts involved in getting the general solution to the differential equation anyway and thus we'll concentrate on this for the remaining illustrations.

Also note that we'll not be demonstrating very much work in solving the characteristic polynomial. Here we are using computational aids and would encourage you to do similar here.  Solving these higher degree polynomials is simply too much work and would obscure the point of these illustrations.