30.2. FIRST ORDER SYSTEMS, CONSTANT COEFFICIENTS 589

2. (AΦ(t)−Φ(t)A)′ = A2Φ(t)−AΦ(t)A = A(AΦ(t)−Φ(t)A) , and also AΦ(0)−Φ(0)A = A−A = 0 for from 1., it follows that AΦ(t)−Φ(t)A = 0.

3. Using 2., and letting t be the variable of differentiation,

(Φ(t +u)−Φ(t)Φ(u))′ = AΦ(t +u)−AΦ(t)Φ(u)

= A(Φ(t +u)−Φ(t)Φ(u))

Also Φ(0+u)−Φ(0)Φ(u) = Φ(u)−Φ(u) = 0 so by part 1., it follows that

Φ(t +u)−Φ(t)Φ(u) = 0.■

30.2.2 Solving a First Order SystemIf you can find the fundamental matrix, it is easy to solve a first order system.

x′ = Ax+f, x(0) = x0 (30.7)

Multiply on the left by Φ(−t) and permute A and Φ(t) as needed using Theorem 30.2.4.

Φ(−t)x′−AΦ(−t)x= Φ(−t)f

(Φ(−t)x)′ = Φ(−t)f (t)

Now integrate and obtain

Φ(−t)x(t)−x0 =∫ t

0Φ(−u)f (u)du

Now multiply on left by Φ(t) to obtain

x(t) = Φ(t)x0 +∫ t

0Φ(t)Φ(−u)f (u)du

= Φ(t)x0 +∫ t

0Φ(t−u)f (u)du

Therefore, there is at most one solution to 30.7 and if there is one, then this is it.

Theorem 30.2.6 There exists a unique solution to 30.7 and it is given by

x(t) = Φ(t)x0 +∫ t

0Φ(t−u)f (u)du

Proof: I just showed there is at most one solution. It only remains to verify that theabove works. However, the formula can be written as

x(t) = Φ(t)x0 +Φ(t)∫ t

0Φ(−u)f (u)du

When t = 0 this yields x0 as it should. Now differentiate. Using the product rule,

x′ (t) = AΦ(t)x0 +AΦ(t)∫ t

0Φ(−u)f (u)du+

I︷ ︸︸ ︷Φ(t)Φ(−t)f (t)

= Ax(t)+f (t) .■