454 CHAPTER 18. LINEAR TRANSFORMATIONS

Theorem 18.6.7 Let f(t) be continuous and let x0 ∈Rp. Then there exists a unique solutionto the equation

x′ = Ax+ f and initial condition x(0) = x0

This solution is given by the formula

x(t) = Φ(t)x0 +Φ(t)∫ t

0Ψ(s) f(s)ds

Where Φ(t) is the fundamental matrix described in Theorem 18.6.5 and Ψ(t) is the funda-mental matrix defined the same way from −A.

Proof: Suppose x′ = Ax+ f. Then x′−Ax = f. Multiply both sides by Ψ(t). Then

(Ψ(t)x)′ = Ψ(t)x′ (t)+Ψ′ (t)x(t)

= Ψ(t)x′ (t)−AΨ(t)x(t)= Ψ(t)x′ (t)−Ψ(t)Ax(t)= Ψ(t)

(x′ (t)−Ax(t)

)Therefore,

(Ψ(t)x)′ = Ψ(t) f(t)

HenceΨ(t)x(t)−x0 =

∫ t

0Ψ(s) f(s)ds

Therefore, multiplying on the left by Φ(t) ,

x(t) = Φ(t)x0 +Φ(t)∫ t

0Ψ(s) f(s)ds

Thus, if there is a solution, this is it. Now if x(t) is given by the above formula, then

x(0) = Φ(0)x0 = x0

and alsox′ (t) = Φ

′ (t)x0 +Φ′ (t)

∫ t

0Ψ(s) f(s)ds+Φ(t)Ψ(t) f(t)

= AΦ(t)x0 +AΦ(t)∫ t

0Ψ(s) f(s)ds+ f(t) = Ax(t)+ f(t) ■

Observation 18.6.8 As a special case when A is a real or complex scalar, the above theo-rem shows that Φ(t)x0 ≡ ∑

∞k=0

tk

k! Akx0 is the solution of the differential equation

x′ (t) = Ax, x(0) = x0.

Another solution to this is eAt where this is defined in Section 1.7. It follows that in thisspecial case, the uniqueness provision of the above theorem shows that

∞

∑k=0

tk

k!Ak = eAt

in the special case that A is a complex number.