6.10. EXERCISES 181

Hint: For the first part write as y′ − ay = 0 and multiply both sides by e−at. Thenexplain why you get

d

dt

(e−aty (t)

)= 0, y (t0) = 0.

Now you finish the argument. To show uniqueness in the second part, suppose

y′ = (a+ ib) y, y (t0) = 0

and verify this requires y (t) = 0. To do this, note

y′ = (a− ib) y, y (t0) = 0

and that |y|2 (t0) = 0 and

d

dt|y (t)|2 = y′ (t) y (t) + y′ (t) y (t)

= (a+ ib) y (t) y (t) + (a− ib) y (t) y (t) = 2a |y (t)|2 .Thus from the first part |y (t)|2 = 0e−2at = 0. Finally observe by a simple computationthat 6.21 is solved by 6.22. For the last part, write the equation as

y′ − ay = f

and multiply both sides by e−at and then integrate from t0 to t using the initialcondition.

38. Now consider A an n×n matrix. By Schur’s theorem there exists unitary Q such that

Q−1AQ = T

where T is upper triangular. Now consider the first order initial value problem

x′ = Ax,x (t0) = x0.

Show there exists a unique solution to this first order system. Hint: Let y = Q−1xand so the system becomes

y′ = Ty,y (t0) = Q−1x0 (6.23)

Now letting y =(y1, · · · , yn)T , the bottom equation becomes

y′n = tnnyn, yn (t0) =(Q−1x0

)n.

Then use the solution you get in this to get the solution to the initial value problemwhich occurs one level up, namely

y′n−1 = t(n−1)(n−1)yn−1 + t(n−1)nyn, yn−1 (t0) =(Q−1x0

)n−1

Continue doing this to obtain a unique solution to 6.23.

39. Now suppose Φ (t) is an n× n matrix of the form

Φ (t) =(

x1 (t) · · · xn (t))

(6.24)

wherex′k (t) = Axk (t) .

Explain whyΦ′ (t) = AΦ (t)

if and only if Φ (t) is given in the form of 6.24. Also explain why if c ∈ Fn,y (t) ≡ Φ (t) csolves the equation y′ (t) = Ay (t) .