(20 pts) Solve the IVP: y'''' - y = u_{1} - u_{2}, y(0) = 0, y'(0) = 0, y''(0) = 0, y'''(0) = 0.

### Solution

For this problem we shall be using the Laplace transform. Because of the vanishing initial conditions, the equation is getting transformed into

(s^{4} - 1) Y(s) = (e^{-s} - e^{-2s})/s.
So that

Y(s) = (e^{-s} - e^{-2s})/s(s^{4} - 1).
Let's ignore the numerator for a moment and find the inverse Laplace transform of G(s) = 1/s(s^{4} - 1). The partial fractions is the way to go. Since s^{4} - 1 = (s^{2} - 1)(s^{2} + 1) = (s + 1)(s - 1)(s^{2} + 1), we should be looking at

1/(s^{4} - 1) = A/(s - 1) + B/(s + 1) + (Cs + D)/(s^{2} + 1) + E/s.
Reduction to the common denominator gives:

1 = A(s + 1)(s^{2} + 1)s + B(s - 1)(s^{2} + 1)s + (Cs + D)(s^{2} - 1)s + E(s^{4} - 1)
Let's compare the coefficients of various powers of s on both sides:

4: 0 = A + B + C +E.

3: 0 = A - B + D.

2: 0 = A + B - C.

1: 0 = A - B - D.

0: 1 = -E

The last equation gives E = -1. From the second and the fourth equations, A - B = 0. So that A = B. From the first and the third equations, A + B = -E/2 = 1/2. Conclude that A = 1/4 and B = 1/4.

Since in the fourth equation A - B = 0, D is also 0. In the third equation C = A + B = 1/2, so that C = 1/2.
G(s) = 1/4·1/(s - 1) + 1/4·1/(s + 1) + 1/2·s/(s^{2} + 1) - 1/s.
This is the Laplace transform of the function

g(t) = 1/4·e^{t} + 1/4·e^{-t} + 1/2·cos(t) - 1.
The solution is the difference of two translations of g:

### Answer

y(t) = u_{1}(t)·g(t - 1) - u_{2}(t)·g(t - 2).