30.6. EXERCISES 611

f (t) F (s) f (t) F (s) f (t) F (s)tneat n!

(s−a)n+1 tn,n ∈ N n!sn+1 eat sinbt b

(s−a)2+b2

eat cosbt s−a(s−a)2+b2 f ∗g(t) F (s)G(s)

Using the table, explain why, for A an n×n matrix, there exists an n×n matrix Φ(t)satisfying

L (Φ)(s) = (sI −A)−1

which has the property that all entries of the kth derivative of Φ(k) (t) have exponentialgrowth. Hint: You should use the formula for the inverse of a matrix in terms of co-factors. The entries of (sI −A)−1 will all be rational functions whose denominatorscan theoretically be factored into products of linear and irreducible quadratics. Thuseach will be the Laplace transform of such a function just described. For the neededtheory of partial fractions, see Problem 40 on Page 46 and the following problem onthat page.

33. ↑Letting Φ(t) be as in the above problem, explain why(I − 1

sA)−1

= Φ(0)+∫

∞

0e−ts

Φ′ (t)dt

Letting s → ∞, explain why Φ(0) = I. Next explain why Φ′ (t) = AΦ(t) . For thislast part, you might show that AΦ(t) and Φ′ (t) have the same Laplace transform.Thus, they will be the same by Theorem 10.3.3. Thus

Φ′ (t) = AΦ(t) ,Φ(0) = I

This is called a fundamental matrix. Show there is at most one solution to the aboveinitial value problem so it is THE fundamental matrix.

34. ↑Show the group property of this fundamental solution, that Φ(t + s) = Φ(t)Φ(s)for any s, t ∈ R. Explain why Φ(−t) = Φ(t)−1. Hint: Use Laplace transforms toshow that if Ψ′ (t) = AΨ(t) ,Ψ(0) = 0, then Ψ(t) = 0. Then consider for s ∈ R,Ψ(t) = Φ(t + s)−Φ(t)Φ(s) .

35. ↑Now show that there is exactly one solution to the initial value problem

x′ (t) = Ax(t)+f (t) , x(0) = x0

and it is given by

x(t) = Φ(t)x0 +Φ(t)∫ t

0Φ(−s)f (s)ds

You just did more than the entire mathematical substance of a typical course in un-dergraduate differential equations other than a few recipes for nonlinear equations.The above formula is called the variation of constants formula or Green’s formula.