Belonging to 1 and 1, respectively, to obtain the solutions

Exercises

In Exercises 1 – 4, find a fundamental matrix solution of the system.

1. 
   
   2
   
   2
   
   y^′ = ¹ x + ³ y;
   x^′ = ³ x + ¹ y
   

2 2
Answer

2. 
   
   
   x^′ = —3x + y y^′ = —5x — y;
   
3. 
   
   —
   
   —
   x^′ = 2x 4y y^′ = 5x 2y;
   

Answer

4. 
   x^′ = —x — y y^′ = x — 3y;
   

In each of Exercises 5 – 8, use the method of variation of constants to find a particular solution of the system of ODEs, and then write down the general solution. Fundamental matrix solutions of the associated homoge- neous systems were found in Exercises 1 – 4.

5. 
   
   
   2
   
   2
   
   y^′ = ¹ x + ³ y + b₂(t),
   x^′ = ³ x + ¹ y + b₁(t)
   

where

1. 
   ^b1 b₂
   
2. 
   ^b1
   

2 2

_et
= _et .
= _et .

b₂

3. 
   ^b1 b₂
   

_6. — —
Answer
—e^t

0
= _tet .

x^′ + 3x y = 3e^—2t

0
y^′ + 5x + y = e^—2t

1. 
   
   ^x′
   

_y′

Answer
2 —4

=
5 —2
_x

_y

₊ —32 cos(4t) _.

2. 
   ^x′
   

_y′
1 1

+
—1 3
_x

_y

b₁(t)

=
b₂(t)
, where

1. 
   
   
   
   b₁(t) b₂(t)
   
2. 
   _b1(t) b₂(t)
   

= e—2t ₁ .

3

0
= _2e—2t .

3. 
   Show that
   

X =
2t² t³
—t² —t³

is a fundamental matrix solution of

—
tx^′ = x 2y ty^′ = x + 4y.
Using variation of constants, find a particular solution of

—
tx^′ = x 2y + b₁(t) ty^′ = x + 4y + b₂(t),

where

1. 
   
   ₂t
   b₁(t) b₂(t)
   

= _t .

2. 
   
   
   b₁(t) b₂(t)
   

_t3

=

.
—t³

Warning: the variation of constants formula is based on the assump- tion that the system has the form →v^′ = A(t)→v +^→b(t); the system in this Exercise must be written in that form before the formula can be used. However, if you use the method of variation of constants there are no worries.
Answer

4. 
   
   Find the general solution of each of the following inhomogeneous systems.
   
   1. 
      
      _(b)
      x^′ = x + y + t^—1 y^′ = —x — y.
      

x^′ = x + y + t^—1

y^′ = —(x + y + t^—1).
_(c) x^′ = 4x + 5y + 5 sin(3t)

(d)
y^′ = —5x — 4y + 3 cos(3t) — 4 sin(3t)
x^′ = sec³(t) — y y^′ = x — sec³(t)

5. 
   
   X
   
   X
   Let C be a constant matrix, and suppose that (t) is a matrix solution of →v^′ = A(t)v. Show that C is also a matrix solution. Answer
   

6. 
   Let X₁(t) and X₂(t) be fundamental matrix solutions of the linear system of differential equations →v^′ = A(t)→v, defined on an interval I. Show that X₁^—1 · X₂ is a constant matrix.
   
7. 
   
   ×
   Let Q be an n n matrix. Show that Q is nonsingular if and only if the columns of q are linearly independent.
   

Answer

1. A Fundamental Matrix Solution of Interest

The Maclaurin series expansion for the function e^at is

_eat
2

a
= 1 + a t + _2! t²
3

a
+ _3! t³
n

a
+ · · · + _n! tⁿ
+ · · · ,

and it converges for all values of t.

Now consider the series with the scalar constant a replaced by a con- stant matrix A. By convention, A⁰ = I is the identity matrix, A¹ = A, A² = A × A, etc.

The first thing to consider is convergence. Let p_ij,k denote the i,jth entry of A^k. Since A⁰ = I, and A¹ = A,

^pij,0 ⁼
1 if i = j _,
0 if i /= j



×
p_ij,1 = a_ij, and so on. Equation (3.25) defines e^At to be the m m matrix whose i,jth entry is
_p ^∞ 1 _k

k!
_ij(t) = _∑ p_ij,k t , (3.26)
k=0
provided that this series is convergent for each i, j.



×
Theorem 3.12 For any m m matrix A, the series (3.26) converges for all i, j and for all values of t.

PROOF. We will show that for any i, j,

_∞

1

∑

k=0 ^k!
_p
tk


(3.27)

| | ≤

ij,k
is convergent. Choose a number B such that for each entry a_ij of A, a_ij B. We’ll see that
|p_ij,k| ≤ m^k—1B^k for k ≥ 1. (3.28)
The proof of (3.28) is by mathematical induction. If k = 1, (3.28) holds because |p_ij,1| = |a_ij| ≤ B. Let k ≥ 2, and assume that (3.28) holds for the

power k — 1; that is


k 2 k 1

|p_ij,k—1| ≤ m ^— B ^— .

Since A^k = A · A^k—1,

^pij,k ^{= a}i1 ^p1j,k—1 ^{+ a}i2 ^p2j,k—1 ^{+ · · · + a}im ^pmj,k—1^.
Starting with the triangle inequality, we have
|p_ij,k | ≤ |a_i1||p_1j,k—1| + |a_i2||p_2j,k—1| + · · · + |a_im ||p_mj,k—1|
≤ B · m^k—2B^k—1 + B · m^k—2B^k—1 + · · · + B · m^k—2B^k—1

₌ `_mk—1 _Bk_.
m te˛r¸ms x

—
Therefore, if (3.28) holds for the power k 1, then it holds for the kth power as well. By the principle of mathematical induction, (3.28) is valid for all

k ≥ 1. Then


k 1 k

1 _k m ^— B _k
k! |p_ij,kt | ≤ _k! |t |,
which implies that the series (3.27) is dominated by the series


^∞ _mk—1 _Bk _k

∑
k=0
_k! |t| .

m
The latter series converges since it is the Maclaurin series expansion of f (t) = ¹ e^mB|t|. By the comparison test, the series (3.27) also converges. This means that the series (3.26) defining p_ij converges absolutely. Since every absolutely convergent series is convergent, the proof is complete.
As a simple example, suppose that Z denotes the zero matrix (all entries equal 0). Then the only nonzero term in the series

2
I + Zt + ¹ Z²t² + · · ·
defining e^Zt is I. It follows that e^Zt = I. The next example is also fairly simple, because the series has only a finite number of nonzero terms.

Example 3.5.1 Let

Calculate e^At.

0 1 0


A = 0 0 1
0 0 0
 .

Solution. By matrix multiplication,

A² =

and A^k = 0 for k ≥ 3. Hence

0 0 1


0 0 0
0 0 0
 ,

e^At = I + At + ¹ (A)²t²


2

= 
1 t ¹ t²

2


0 1 t
0 0 1

Example 3.5.2 Calculate e^At, where

0 2
A = ^{1 0} .

Solution. If
_{B =} a 0
and C = ^{c 0}

0 b

0 d
are diagonal matrices, you can readily verify that

0 bd
BC = ^{ac 0} .



A =

.
Thus, the product of two diagonal matrices is also a diagonal matrix. The diagonal entries of the product matrix are obtained simply by multiplying the respective diagonal entries of B and C. It follows that we can find the nth power of a diagonal matrix by raising each diagonal entry to the nth power:

Thus
_n 1ⁿ 0
0 2ⁿ

_eAt
^∞ 1 _{k k}

k!
= _∑ A t
k=0

" ∑_{∞ tk} 0 #

=
^k=0 k!

⁰ ∑
∞

=

.
k=0
(2t)^k

k!

e^t 0
_{0 e}2t

The matrix exponential shares many properties with the ordinary expo- nential function, but there are limitations. The proof that e^r+s = e^re^s does not emphasize the commutative law, rs = sr, but it uses that fact. Multipli- cation of square matrices is not always commutative, so the matrix version

of that identity is as follows.