1. (20 pts) Solve the IVP: y'''' - y = u1 - u2, 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

(s4 - 1) Y(s) = (e-s - e-2s)/s.

So that

Y(s) = (e-s - e-2s)/s(s4 - 1).

Let's ignore the numerator for a moment and find the inverse Laplace transform of G(s) = 1/s(s4 - 1). The partial fractions is the way to go. Since s4 - 1 = (s2 - 1)(s2 + 1) = (s + 1)(s - 1)(s2 + 1), we should be looking at

1/(s4 - 1) = A/(s - 1) + B/(s + 1) + (Cs + D)/(s2 + 1) + E/s.

Reduction to the common denominator gives:

1 = A(s + 1)(s2 + 1)s + B(s - 1)(s2 + 1)s + (Cs + D)(s2 - 1)s + E(s4 - 1)

Let's compare the coefficients of various powers of s on both sides:

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/(s2 + 1) - 1/s.

This is the Laplace transform of the function

g(t) = 1/4·et + 1/4·e-t + 1/2·cos(t) - 1.

The solution is the difference of two translations of g:

Answer

y(t) = u1(t)·g(t - 1) - u2(t)·g(t - 2).