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

3.4. MATRIX SOLUTIONS AND VARIATION OF CONSTANTS

^—
belonging to 1 and 1, respectively, to obtain the solutions →v₁(t) = e^t→e₁
and →v₂(t) = e^—t→e₂. Since
→v (t) = ^et and v→ (t) = ^e—t ,

1. 
   
   X (t) =
   
   .
   _et
   
2. 
   _—e—t
   

is a matrix solution.
_et _e—t
_et _—e—t

1 —1
Then X (0) = ^{1 1} ; since det X (0) = —2 if follows from
proposition 3.4.3 that X (t) is a fundamental matrix solution.



Example 3.4.2 Find the general solution of the system 3.21, and also the solution

5
with initial value →v(0) = ³ .



Solution. In example 3.4.1, we found a fundamental matrix solution,

The general solution is

→v =
=
_et _e—t

X =
_et _—e—t



X t
→c
e e^—t c₁
t

=
e —e^—t c₂
c₁e^t + c₂e^—t c₁e^t — c₂e^—t

1

—1
= c₁e^{t 1} + c₂e^{—t 1} .



X
To find the solution of the IVP, we must find a vector →c such that (0)→c is equal to the specified value of →v(0). Thus we must solve the equation

1 1
1 —1
_c1 = ₃ .

5

c₂
or, equivalently, (c₁, c₂) must satisfy the system
c₁ + c₂ = 3
c₁ — c₂ = 5

—1
which yields (c₁, c₂) = (4, —1). The solution of the IVP is therefore

1
→v(t) = 4e^{t 1}
— e—t ₁

Inhomogeneous systems

The method for solving an inhomogeneous system of linear ODEs,
→v^′ = A(t)→v + ^→f (t), (3.22)

×
where A(t) is an n n matrix, is basically the same as the method we used in section 1.3 to solve a single first order linear ODE. We start with the general solution of the associated homogeneous system,
→v^′ = A(t)→v,
which we can express in terms of a fundamental matrix solution,
→v_h(t) = X (t)→c
and add a particular solution →v_p(t) of the inhomogeneous equation.

X
Theorem 3.10 Suppose that all entries of the coefficient matrix A(t) and source vector ^→f (t) in (3.22) are continuous on an interval (a, b). Let →v_p(t) be a particular solution of the system (3.22), and let (t) be a fundamental matrix solution of the associated homogeneous system. Then the general solution of the system (3.22) on (a, b) is
→v(t) = →v_p(t) + X (t)→c.
PROOF. Let →v(t) be an arbitrary solution of the inhomogeneous sys- tem (3.22), and put →y(t) = →v(t) — →v_p(t). Then
→y^′ = →v^′ — →v^′_p
= [A(t)→v + ^→f (t)] — [A(t)→v_p + ^→f (t)]
= A(t)→v — A(t)→v_p = A→y.
It follows that →y(t) is a solution of the associated homogeneous system. Therefore there is a unique constant vector →c such that
→y(t) = X (t)→c;

X
and it follows that →v(t) = →v_p(t) + (t)→c.

X
Conversely, suppose that →v(t) = →v_p(t) + (t)→c, where →c is a constant vector. Then

dt ^p
d (→v(t)) = →v^′ (t) + X ^′(t)→c
= A(t)→v_p(t) + ^→f (t) + A(t)X (t)→c
= A(t)(→v_p(t) + X (t)→c) + ^→f (t)
= A(t)→v(t) + ^→f (t);
in other words, →v(t) is a solution of (3.22).
Theorem 3.10 can best be deployed in conjunction with a method for
finding particular solutions of systems.

Variation of constants

The method of variation of constants is used to find a particular solution of an inhomogeneous system. Recall from section 1.3, where the method was introduced for the scalar case, that the general solution of the associated homogeneous equation was a required input. For systems, it is convenient to start with a fundamental matrix solution of the associated homogeneous system.
The key to adapting the method for systems of linear ODEs is inversion of a matrix solution. This replaces division by y_h(t) in the scalar case (as in equation (1.11)).

P
Let P be a square matrix. A square matrix Q is called the inverse matrix of if P Q = Q P = I, where I is the identity matrix. We will use the customary notation P^—1 for the inverse matrix of P. The following theorem is from linear algebra. Although it is true for square matrices of any size, our proof only works for 2 × 2 matrices.

Theorem 3.11 Let

b d
_{P =} a c
be a nonsingular matrix. Then P has an inverse matrix.

—
PROOF By proposition 3.2.2, det(P) = ad bc is equal to 0 if and only if
P is singular.
Thus, if P is nonsingular, det(P) /= 0. Define a matrix Q as
_{Q =} ¹ d —b _.
det(P) ^{—c a}


You can verify that P Q = I and Q P = I by matrix multiplication.



X
We will now see how to find a particular solution of an inhomogeneous system of ODEs. Let (t) be a fundamental matrix solution of a homo- geneous system →v^′ = A(t)→v. To find a particular solution of an inhomoge- neous system

→v^′ = A(t)→v + ^→f (t) (3.23)

set


→v_p(t) = X (t)w→ (t),

where w→ (t) is a vector function that will be determined. By the product rule for differentiation,

→v^′_p = X ^′(t)w→ (t) + X (t)w→ ^′(t)
= A(t)X (t)w→ (t) + X (t)w→ ^′(t)
= A(t)→v_p(t) + X (t)w→ ^′(t).
It follows that →v_p(t) is a particular solution of the system (3.23) if and only if
X (t)w→ ^′(t) = ^→f (t). (3.24)

X X
Since (t) is nonsingular, it has an inverse ^—1(t). Multiplying (3.24) by this, we have

If t₀ ∈ (a, b), we can put

w→ ^′(t) = X ^—1(t)^→f (t).

∫ X
w→ (t) = t ^—1(s)^→f (s) ds.
t₀

Now that w→ (t) has been determined we can find →v_p(t) by multiplying by

X (t). This can be summarized as follows:

It is not necessary to memorize this formula; you just need to remember to substitute →v = X (t)w→ in (3.23).

Example 3.4.3 Find a particular solution of

_x1′ _x2′
= x₂ e^—t

—
= x₁ + e^—t

Solution. In example 3.4.1, it was shown that X (t) = ^{e e}

_et _—e—t
t —t
is a

=



=
fundamental matrix solution of the associated homogeneous system (3.21). Set →v_p(t) = X (t)w→ (t). Thus

x₁ e^t e^—t w₁
x₂ e^t —e^—t w₂
e^tw₁ + e^—tw₂ e^tw₁ — e^—tw₂

=

+ _—
Differentiating, and using the product rule we have

_x1′ _x2′
_et_w1′ _{+ e}—t_w2′ _et_w1′ _{— e}—t_w2′
e^tw₁ e^—tw₂ e^tw₁ + e^—tw₂

= <sub>—

e</sub>—t
and the right side of our system is

_{—
x2 e}—t x₁ + e^—t
e^tw₁ e^—tw₂ e^tw₁ + e^—tw₂
+ _—e—t

After canceling, we have


_et_w1′ _{+ e}—t_w2′ _et_w1′ _{— e}—t_w2′


= —e^—t
= e^—t.

The solution of this equation is w₁^′ = 0, w₂^′ = —1, or

—1
w→ ^′(t) = ⁰ .



—t

(t) =



= _—

.
Thus w→ (t) = ⁰ , and we have the particular solution

→v_p
et _e—t ₀
e^t —e^—t —t
_te—t _te—t

X
The general solution is →v_p(t) + →v_h(t), where →v_h(t) = (t)→c denotes the general solution of the associated homogeneous equation.