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8.2 Nonhomogeneous Linear Systems

In Section 5.5 we exhibited two techniques for finding a single particular solution of a single nonhomogeneous nth-order linear differential equation—the method of undetermined coefficients and the method of variation of parameters. Each of these may be generalized to nonhomogeneous linear systems. In a linear system modeling a physical situation, nonhomogeneous terms typically correspond to external influences, such as the inflow of liquid to a cascade of brine tanks or an external force acting on a mass-and-spring system.

Given the nonhomogeneous first-order linear system

x' = A x + f (t)

$x^{′} = Ax + f (t)$ (1)

where A is an n×n $n \times n$ constant matrix and the “nonhomogeneous term” f(t) is a given continuous vector-valued function, we know from Theorem 4 of Section 7.2 that a general solution of Eq. (1) has the form

x (t) = x c (t) + x p (t),

$x (t) = x_{c} (t) + x_{p} (t),$ (2)

where

xc(t)=c1x1(t)+c2x2(t)+⋯+cnxn(t) $x_{c} (t) = c_{1} x_{1} (t) + c_{2} x_{2} (t) + \dots + c_{n} x_{n} (t)$ is a general solution of the associated homogeneous system x'=Ax $x ′ = Ax$ , and
xp(t) $x_{p} (t)$ is a single particular solution of the original nonhomogeneous system in (1).

Preceding sections have dealt with xc(t) $x_{c} (t)$ , so our task now is to find xp(t) $x_{p} (t)$ .

Undetermined Coefficients

First we suppose that the nonhomogeneous term f(t) in (1) is a linear combination (with constant vector coefficients) of products of polynomials, exponential functions, and sines and cosines. Then the method of undetermined coefficients for systems is essentially the same as for a single linear differential equation. We make an intelligent guess as to the general form of a particular solution xp $x_{p}$ , then attempt to determine the coefficients in xp $x_{p}$ by substitution in Eq. (1). Moreover, the choice of this general form is essentially the same as in the case of a single equation (discussed in Section 5.5); we modify it only by using undetermined vector coefficients rather than undetermined scalars. We will therefore confine the present discussion to illustrative examples.

Example 1

Find a particular solution of the nonhomogeneous system

x' = [3725] x + [3 2 t] .

$x^{′} = [\begin{array}{r} 3 & 2 \\ 7 & 5 \end{array}] x + [\begin{array}{r} 3 \\ 2 t \end{array}] .$ (3)

Solution

The nonhomogeneous term f=[32t]T $f = {[\begin{array}{r} 3 & 2 t \end{array}]}^{T}$ is linear, so it is reasonable to select a linear trial particular solution of the form

x p (t) = a t + b = [a 1 a 2] t + [b 1 b 2] .

$x_{p} (t) = a t + b = [\begin{array}{r} a_{1} \\ a_{2} \end{array}] t + [\begin{array}{r} b_{1} \\ b_{2} \end{array}] .$ (4)

Upon substitution of x=xp $x = x_{p}$ in Eq. (3), we get

[a 1 a 2] = = [3725] [a 1 t + b 2 a 2 t + b 2] + [3 2 t] [3 a 1 + 2 a 2 7 a 1 + 5 a 2 + 2] t + [3 b 1 + 2 b 2 + 3 7 b 1 + 5 b 2] .

$\begin{array}{rcl} [\begin{array}{r} a_{1} \\ a_{2} \end{array}] & = & [\begin{array}{r} 3 & 2 \\ 7 & 5 \end{array}] [\begin{array}{r} a_{1} t + b_{2} \\ a_{2} t + b_{2} \end{array}] + [\begin{array}{r} 3 \\ 2 t \end{array}] \\ = & [\begin{array}{c} 3 a_{1} + 2 a_{2} \\ 7 a_{1} + 5 a_{2} + 2 \end{array}] t + [\begin{array}{c} 3 b_{1} + 2 b_{2} + 3 \\ 7 b_{1} + 5 b_{2} \end{array}] . \end{array}$

We equate the coefficients of t and the constant terms (in both x1 $x_{1}$ - and x2 $x_{2}$ -components) and thereby obtain the equations

3 a 1 + 2 a 2 7 a 1 + 5 a 2 + 2 3 b 1 + 2 b 2 + 3 7 b 1 + 5 b 2 = = = = 0, 0, a 1, a 2,

$\begin{array}{rcl} 3 a_{1} + 2 a_{2} & = & 0, \\ 7 a_{1} + 5 a_{2} + 2 & = & 0, \\ 3 b_{1} + 2 b_{2} + 3 & = & a_{1}, \\ 7 b_{1} + 5 b_{2} & = & a_{2}, \end{array}$ (5)

We solve the first two equations in (5) for a1=4 $a_{1} = 4$ and a2=−6 $a_{2} = - 6$ . With these values we can then solve the last two equations in (5) for b1=17 $b_{1} = 17$ and b2=−25 $b_{2} = - 25$ . Substitution of these coefficients in Eq. (4) gives the particular solution x=[x1x2]T $x = {[\begin{array}{r} x_{1} & x_{2} \end{array}]}^{T}$ of (3) described in scalar form by

x 1 (t) x 2 (t) = = 4 t + 17, - 6 t - 25.

$\begin{array}{rcr} x_{1} (t) & = & 4 t + 17, \\ x_{2} (t) & = & - 6 t - 25. \end{array}$

Example 2

Cascading brine tanks Figure 8.2.1 shows the system of three brine tanks investigated in Example 2 of Section 7.3. The volumes of the three tanks are V1=20,=V2=40, $V_{1} = 20, = V_{2} = 40,$ and V3=50(gal), $V_{3} = 50 (gal),$ and the common flow rate is r=10 $r = 10$ (gal/min). Suppose that all three tanks contain fresh water initially, but that the inflow to tank 1 is brine containing 2 pounds of salt per gallon, so that 20 pounds of salt flow into tank 1 per minute. Referring to Eq. (18) in Section 7.3, we see that the vector x(t)=[x1(t)x2(t)x3(t)]T $x (t) = {[\begin{matrix} x_{1} (t) & x_{2} (t) & x_{3} (t) \end{matrix}]}^{T}$ of amounts of salt (in pounds) in the three tanks at time t satisfies the nonhomogeneous initial value problem

d x d t = ⎡ ⎣ ⎢ - 0.5 0.5 0 0 - 0.25 0.25 00 - 0.2 ⎤ ⎦ ⎥ x + ⎡ ⎣ ⎢ 2000 ⎤ ⎦ ⎥, x (0) = ⎡ ⎣ ⎢ 000 ⎤ ⎦ ⎥ .

$\frac{d x}{d t} = [\begin{array}{l} - 0.5 & 0 & 0 \\ 0.5 & - 0.25 & 0 \\ 0 & 0.25 & - 0.2 \end{array}] x + [\begin{array}{r} 20 \\ 0 \\ 0 \end{array}], x (0) = [\begin{array}{l} 0 \\ 0 \\ 0 \end{array}] .$ (6)

FIGURE 8.2.1.

The three brine tanks of Example 2.

The nonhomogeneous term f=[2000]T $f = {[\begin{array}{r} 20 & 0 & 0 \end{array}]}^{T}$ here corresponds to the 20 lb/min inflow of salt to tank 1, with no (external) inflow of salt into tanks 2 and 3.

Because the nonhomogeneous term is constant, we naturally select a constant trial function xp=[a1a2a3]T $x_{p} = {[\begin{array}{r} a_{1} & a_{2} & a_{3} \end{array}]}^{T}$ , for which x'p≡0. $x_{p}^{′} \equiv 0 .$ Then substitution of x=xp $x = x_{p}$ in (6) yields the system

⎡ ⎣ ⎢ 000 ⎤ ⎦ ⎥ = ⎡ ⎣ ⎢ - 0.5 0.5 0 0 - 0.25 0.25 00 - 0.2 ⎤ ⎦ ⎥ ⎡ ⎣ ⎢ a 1 a 2 a 3 ⎤ ⎦ ⎥ + ⎡ ⎣ ⎢ 2000 ⎤ ⎦ ⎥

$[\begin{array}{r} 0 \\ 0 \\ 0 \end{array}] = [\begin{array}{l} - 0.5 & 0 & 0 \\ 0.5 & - 0.25 & 0 \\ 0 & 0.25 & - 0.2 \end{array}] [\begin{array}{r} a_{1} \\ a_{2} \\ a_{3} \end{array}] + [\begin{array}{r} 20 \\ 0 \\ 0 \end{array}]$

that we readily solve for a1=40, a2=80, $a_{1} = 40, a_{2} = 80,$ and a3=100 $a_{3} = 100$ in turn. Thus our particular solution is xp(t)=[4080100]T $x_{p} (t) = {[\begin{array}{r} \begin{array}{l} 40 & 80 & 100 \end{array} \end{array}]}^{T}$ .

In Example 2 of Section 7.3 we found the general solution

x c (t) = c 1 ⎡ ⎣ ⎢ 3 - 6 5 ⎤ ⎦ ⎥ e - t / 2 + c 2 ⎡ ⎣ ⎢ 01 - 5 ⎤ ⎦ ⎥ e - t / 4 + c 3 ⎡ ⎣ ⎢ 001 ⎤ ⎦ ⎥ e - t / 5

$x_{c} (t) = c_{1} [\begin{array}{r} 3 \\ - 6 \\ 5 \end{array}] e^{- t / 2} + c_{2} [\begin{array}{r} 0 \\ 1 \\ - 5 \end{array}] e^{- t / 4} + c_{3} [\begin{array}{r} 0 \\ 0 \\ 1 \end{array}] e^{- t / 5}$

of the associated homogeneous system, so a general solution x=xc+xp $x = x_{c} + x_{p}$ of the nonhomogeneous system in (6) is given by

x (t) = c 1 ⎡ ⎣ ⎢ 3 - 6 5 ⎤ ⎦ ⎥ e - t / 2 + c 2 ⎡ ⎣ ⎢ 01 - 5 ⎤ ⎦ ⎥ e - t / 4 + c 3 ⎡ ⎣ ⎢ 001 ⎤ ⎦ ⎥ e - t / 5 + ⎡ ⎣ ⎢ 4080100 ⎤ ⎦ ⎥ .

$x (t) = c_{1} [\begin{array}{r} 3 \\ - 6 \\ 5 \end{array}] e^{- t / 2} + c_{2} [\begin{array}{r} 0 \\ 1 \\ - 5 \end{array}] e^{- t / 4} + c_{3} [\begin{array}{r} 0 \\ 0 \\ 1 \end{array}] e^{- t / 5} + [\begin{array}{r} 40 \\ 80 \\ 100 \end{array}] .$ (7)

When we apply the zero initial conditions in (6), we get the scalar equations

3 c 1 - 6 c 1 5 c 1 + - c 2 5 c 2 + c 3 + + + 4080100 = = = 0, 0, 0

$\begin{array}{rcrcrcrcl} 3 c_{1} & + & 40 & = & 0, \\ - 6 c_{1} & + & c_{2} & + & 80 & = & 0, \\ 5 c_{1} & - & 5 c_{2} & + & c_{3} & + & 100 & = & 0 \end{array}$

that are readily solved for c1=−403, c2=−160 $c_{1} = - \frac{40}{3}, c_{2} = - 160$ , and c3=−25003 $c_{3} = - \frac{2500}{3}$ . Substituting these coefficients in Eq. (7), we find that the amounts of salt in the three tanks at time t are given by

x 1 (t) x 2 (t) x 3 (t) = = = 40 - 40 e - t / 2, 80 + 80 e - t / 2 - 160 e - t / 4 100 + 100 3 (- 2 e - t / 2 + 24 e - t / 4 - 25 e - t / 5) .

$\begin{array}{rcl} x_{1} (t) & = & 40 - 40 e^{- t / 2}, \\ x_{2} (t) & = & 80 + 80 e^{- t / 2} - 160 e^{- t / 4} \\ x_{3} (t) & = & 100 + \frac{100}{3} (- 2 e^{- t / 2} + 24 e^{- t / 4} - 25 e^{- t / 5}) . \end{array}$ (8)

As illustrated in Fig. 8.2.2, we see the salt in each of the three tanks approaching, as t→+∞ $t \to + \infty$ , a uniform density of 2 lb/gal—the same as the salt density in the inflow to tank 1.

FIGURE 8.2.2.

Solution curves for the amount of salt defined in (8).

In the case of duplicate expressions in the complementary function and the nonhomogeneous terms, there is one difference between the method of undetermined coefficients for systems and for single equations (Rule 2 in Section 5.5). For a system, the usual first choice for a trial solution must be multiplied not only by the smallest integral power of t that will eliminate duplication, but also by all lower (nonnegative integral) powers of t as well, and all the resulting terms must be included in the trial solution.

Example 3

Consider the nonhomogeneous system

x' = [43 2 - 1] x - [154] t e - 2 t .

$x^{′} = [\begin{array}{r} 4 & 2 \\ 3 & - 1 \end{array}] x - [\begin{array}{r} 15 \\ 4 \end{array}] t e^{- 2 t} .$ (9)

In Example 1 of Section 7.3 we found the solution

x c (t) = c 1 [1 - 3] e - 2 t + c 2 [21] e 5 t

$x_{c} (t) = c_{1} [\begin{array}{r} 1 \\ - 3 \end{array}] e^{- 2 t} + c_{2} [\begin{array}{r} 2 \\ 1 \end{array}] e^{5 t}$ (10)

of the associated homogeneous system. A preliminary trial solution xp(t)=ate−2t+be−2t $x_{p} (t) = a t e^{- 2 t} + b e^{- 2 t}$ exhibits duplication with the complementary function in (10). We would therefore select

x p (t) = a t 2 e - 2 t + b t e - 2 t + c e - 2 t

$x_{p} (t) = a t^{2} e^{- 2 t} + b t e^{- 2 t} + c e^{- 2 t}$

as our trial solution, and we would then have six scalar coefficients to determine. It is simpler to use the method of variation of parameters, our next topic.

Variation of Parameters

Recall from Section 5.5 that the method of variation of parameters may be applied to a linear differential equation with variable coefficients and is not restricted to nonhomogeneous terms involving only polynomials, exponentials, and sinusoidal functions. The method of variation of parameters for systems enjoys the same flexibility and has a concise matrix formulation that is convenient for both practical and theoretical purposes.

We want to find a particular solution xp $x_{p}$ of the nonhomogeneous linear system

x' = P (t) x + f (t),

$x^{′} = P (t) x + f (t),$ (11)

given that we have already found a general solution

x c (t) = c 1 x 1 (t) + c 2 x 2 (t) + \dots + c n x n (t)

$x_{c} (t) = c_{1} x_{1} (t) + c_{2} x_{2} (t) + \dots + c_{n} x_{n} (t)$ (12)

of the associated homogeneous system

x' = P (t) x .

$x^{′} = P (t) x .$ (13)

We first use the fundamental matrix Φ(t) $Φ (t)$ with column vectors x1, x2, …, $x_{1}, x_{2}, \dots,$ xn $x_{n}$ to rewrite the complementary function in (12) as

x c (t) = Φ (t) c,

$x_{c} (t) = Φ (t) c,$ (14)

where c denotes the column vector whose entries are the coefficients c1,c2,…,cn $c_{1}, c_{2}, \dots, c_{n}$ . Our idea is to replace the vector “parameter” c with a variable vector u(t). Thus we seek a particular solution of the form

x p (t) = Φ (t) u (t) .

$x_{p} (t) = Φ (t) u (t) .$ (15)

We must determine u(t) so that xp $x_{p}$ does, indeed, satisfy Eq. (11).

The derivative of xp(t) $x_{p} (t)$ is (by the product rule)

x' p (t) = Φ' (t) u (t) + Φ (t) u' (t) .

$x_{p}^{′} (t) = Φ ′ (t) u (t) + Φ (t) u ′ (t) .$ (16)

Hence substitution of Eqs. (15) and (16) in (11) yields

Φ' (t) u (t) + Φ (t) u' (t) = P (t) Φ (t) u (t) + f (t) .

$Φ ′ (t) u (t) + Φ (t) u ′ (t) = P (t) Φ (t) u (t) + f (t) .$ (17)

But

Φ' (t) = P (t) Φ (t)

$Φ ′ (t) = P (t) Φ (t)$ (18)

because each column vector of Φ(t) $Φ (t)$ satisfies Eq. (13). Therefore, Eq. (17) reduces to

Φ (t) u' (t) = f (t) .

$Φ (t) u ′ (t) = f (t) .$ (19)

Thus it suffices to choose u(t) so that

u' (t) = Φ (t) - 1 f (t);

$u ′ (t) = Φ {(t)}^{- 1} f (t);$ (20)

that is, so that

u (t) = \int Φ (t) - 1 f (t) d t .

$u (t) = \int Φ {(t)}^{- 1} f (t) d t .$ (21)

Upon substitution of (21) in (15), we finally obtain the desired particular solution, as stated in the following theorem.

This is the variation of parameters formula for first-order linear systems. If we add this particular solution and the complementary function in (14), we get the general solution

x (t) = Φ (t) c + Φ (t) \int Φ (t) - 1 f (t) d t

$x (t) = Φ (t) c + Φ (t) \int Φ {(t)}^{- 1} f (t) d t$ (23)

of the nonhomogeneous system in (11).

The choice of the constant of integration in Eq. (22) is immaterial, for we need only a single particular solution. In solving initial value problems it often is convenient to choose the constant of integration so that xp(a)=0 $x_{p} (a) = 0$ , and thus integrate from a to t:

x p (t) = Φ (t) \int t a Φ (s) - 1 f (s) d s .

$x_{p} (t) = Φ (t) \int_{a}^{t} Φ {(s)}^{- 1} f (s) d s .$ (24)

If we add the particular solution of the nonhomogeneous problem

x' = P (t) x + f (t), x (a) = 0

$x^{′} = P (t) x + f (t), x (a) = 0$

in (24) to the solution xc(t)=Φ(t)Φ(a)−1xa $x_{c} (t) = Φ (t) Φ {(a)}^{- 1} x_{a}$ of the associated homogeneous problem x'=P(t)x $x ′ = P (t) x$ , x(a)=xa $x (a) = x_{a}$ , we get the solution

x (t) = Φ (t) Φ (a) - 1 x a + Φ (t) \int t a Φ (s) - 1 f (s) d s

$x (t) = Φ (t) Φ {(a)}^{- 1} x_{a} + Φ (t) \int_{a}^{t} Φ {(s)}^{- 1} f (s) d s$ (25)

of the nonhomogeneous initial value problem

x' = P (t) x + f (t), x (a) = x a .

$x^{′} = P (t) x + f (t), x (a) = x_{a} .$ (26)

Equations (22) and (25) hold for any fundamental matrix Φ(t) $Φ (t)$ of the homogeneous system x'=P(t)x $x ′ = P (t) x$ . In the constant-coefficient case P(t)≡A $P (t) \equiv A$ we can use for Φ(t) $Φ (t)$ the exponential matrix eAt $e^{A t}$ —that is, the particular fundamental matrix such that Φ(0)=I $Φ (0) = I$ . Then, because (eAt)−1=e−At ${(e^{A t})}^{- 1} = e^{- A t}$ , substitution of Φ(t)=eAt $Φ (t) = e^{A t}$ in (22) yields the particular solution

x p (t) = e A t \int e - A t f (t) d t

$x_{p} (t) = e^{A t} \int e^{- A t} f (t) d t$ (27)

of the nonhomogeneous system x'=P(t)x+f(t) $x ′ = P (t) x + f (t)$ . Similarly, substitution of Φ(t)=eAt $Φ (t) = e^{A t}$ in Eq. (25) with a=0 $a = 0$ yields the solution

x (t) = e A t x 0 + e A t \int t o e - A t f (t) d t

$x (t) = e^{A t} x_{0} + e^{A t} \int_{o}^{t} e^{- A t} f (t) d t$ (28)

of the initial value problem

x' = P (t) x + f (t), x (0) = x 0 .

$x^{′} = P (t) x + f (t), x (0) = x_{0.}$ (29)

Remark

If we retain t as the independent variable but use s for the variable of integration, then the solutions in (27) and (28) can be rewritten in the forms

x p (t) = \int e - A (s - t) f (s) d s and x (t) = e A t x 0 + \int t 0 e - A (s - t) f (s) d s .

$x_{p} (t) = \int e^{- A (s - t)} f (s) d s and x (t) = e^{A t} x_{0} + \int_{0}^{t} e^{- A (s - t)} f (s) d s .$

Example 4

Solve the initial value problem

x' = [43 2 - 1] x - [154] t e - 2 t, x (0) = [73] .

$x^{′} = [\begin{array}{r} 4 & 2 \\ 3 & - 1 \end{array}] x - [\begin{array}{r} 15 \\ 4 \end{array}] t e^{- 2 t}, x (0) = [\begin{array}{r} 7 \\ 3 \end{array}] .$ (30)

Solution

The solution of the associated homogeneous system is displayed in Eq. (10). It gives the fundamental matrix

Φ (t) = [e - 2 t - 3 e - 2 t 2 e 5 t e 5 t] with Φ (0) - 1 = 1 7 [13 - 2 1] .

$Φ (t) = [\begin{array}{r} e^{- 2 t} & 2 e^{5 t} \\ - 3 e^{- 2 t} & e^{5 t} \end{array}] with Φ {(0)}^{- 1} = \frac{1}{7} [\begin{array}{r} 1 & - 2 \\ 3 & 1 \end{array}] .$

It follows by Eq. (28) in Section 8.1 that the matrix exponential for the coefficient matrix A in (30) is

e A t = Φ (t) Φ (0) - 1 = = [e - 2 t - 3 e - 2 t 2 e 5 t e 5 t] \cdot 1 7 [13 - 2 1] 1 7 [e - 2 t + 6 e 5 t - 3 e - 2 t + 3 e 5 t - 2 e - 2 t + 2 e 5 t 6 e - 2 t + e 5 t] .

$\begin{array}{rcl} e^{A t} = Φ (t) Φ {(0)}^{- 1} & = & [\begin{array}{r} e^{- 2 t} & 2 e^{5 t} \\ - 3 e^{- 2 t} & e^{5 t} \end{array}] \cdot \frac{1}{7} [\begin{array}{r} 1 & - 2 \\ 3 & 1 \end{array}] \\ = & \frac{1}{7} [\begin{array}{r} e^{- 2 t} + 6 e^{5 t} & - 2 e^{- 2 t} + 2 e^{5 t} \\ - 3 e^{- 2 t} + 3 e^{5 t} & 6 e^{- 2 t} + e^{5 t} \end{array}] . \end{array}$

Then the variation of parameters formula in Eq. (28) gives

e - A t x (t) = = = = x 0 + \int t 0 e - A s f (s) d s [73] + \int t 0 1 7 [e 2 s + 6 e - 5 s - 3 2 s + 3 e - 5 s - 2 e 2 s + 2 e - 5 s 6 e 2 s + e - 5 s] [- 15 s e - 2 s - 4 s e - 2 s] d s [73] + \int t 0 [- s - 14 s e - 7 s 3 s - 7 s e - 7 s] d s [73] + 1 14 [- 4 - 7 t 2 + 4 e - 7 t + 28 t e - 7 t - 2 + 21 t 2 + 2 e - 7 t + 14 t e - 7 t] .

$\begin{array}{rcl} e^{- A t} x (t) & = & x_{0} + \int_{0}^{t} e^{- A s} f (s) d s \\ = & [\begin{array}{r} 7 \\ 3 \end{array}] + \int_{0}^{t} \frac{1}{7} [\begin{array}{r} e^{2 s} + 6 e^{- 5 s} & - 2 e^{2 s} + 2 e^{- 5 s} \\ - 3^{2 s} + 3 e^{- 5 s} & 6 e^{2 s} + e^{- 5 s} \end{array}] [\begin{array}{r} - 15 s e^{- 2 s} \\ - 4 s e^{- 2 s} \end{array}] d s \\ = & [\begin{array}{r} 7 \\ 3 \end{array}] + \int_{0}^{t} [\begin{array}{r} - s - 14 s e^{- 7 s} \\ 3 s - 7 s e^{- 7 s} \end{array}] d s \\ = & [\begin{array}{r} 7 \\ 3 \end{array}] + \frac{1}{14} [\begin{array}{r} - 4 - 7 t^{2} + 4 e^{- 7 t} + 28 t e^{- 7 t} \\ - 2 + 21 t^{2} + 2 e^{- 7 t} + 14 t e^{- 7 t} \end{array}] . \end{array}$

Therefore,

e - A t x (t) = 1 14 [94 - 7 t 2 + 4 e - 7 t + 28 t e - 7 t 40 + 21 t 2 + 2 e - 7 t + 14 t e - 7 t] .

$e^{- A t} x (t) = \frac{1}{14} [\begin{array}{r} 94 - 7 t^{2} + 4 e^{- 7 t} + 28 t e^{- 7 t} \\ 40 + 21 t^{2} + 2 e^{- 7 t} + 14 t e^{- 7 t} \end{array}] .$

Upon multiplication of the right-hand side here by eAt $e^{A t}$ , we find that the solution of the initial value problem in (30) is given by

x (t) = = 1 7 [e - 2 t + 6 e 5 t - 3 e - 2 t + 3 e 5 t - 2 e - 2 t + 2 e 5 t 6 e - 2 t + e 5 t] \cdot 1 14 [94 - 7 t 2 + 4 e - 7 t + 28 t e - 7 t 40 + 21 t 2 + 2 e - 7 t + 14 t e - 7 t] 1 14 [(6 + 28 t - 7 t 2) e - 2 t + 92 e 5 t (- 4 + 14 t + 21 t 2) e - 2 t + 46 e 5 t] .

$\begin{array}{rcl} x (t) & = & \frac{1}{7} [\begin{array}{r} e^{- 2 t} + 6 e^{5 t} & - 2 e^{- 2 t} + 2 e^{5 t} \\ - 3 e^{- 2 t} + 3 e^{5 t} & 6 e^{- 2 t} + e^{5 t} \end{array}] \cdot \frac{1}{14} [\begin{array}{r} 94 - 7 t^{2} + 4 e^{- 7 t} + 28 t e^{- 7 t} \\ 40 + 21 t^{2} + 2 e^{- 7 t} + 14 t e^{- 7 t} \end{array}] \\ = & \frac{1}{14} [\begin{array}{r} (6 + 28 t - 7 t^{2}) e^{- 2 t} + 92 e^{5 t} \\ (- 4 + 14 t + 21 t^{2}) e^{- 2 t} + 46 e^{5 t} \end{array}] . \end{array}$

In conclusion, let us investigate how the variation of parameters formula in (22) “reconciles” with the variation of parameters formula in Theorem 1 of Section 5.5 for the second-order linear differential equation

y'' + P y' + Q y = f (t) .

$y^{″} + P y^{′} + Q y = f (t) .$ (31)

If we write y=x1, y'=x'1=x2, y''=x′′1=x'2 $y = x_{1}, y ′ = x_{1}^{′} = x_{2}, y ″ = x_{1}^{″} = x_{2}^{′}$ , then the single equation in (31) is equivalent to the linear system x'1=x2, x'2=−Qx1−Px2+f(t), $x_{1}^{′} = x_{2}, x_{2}^{′} = - Q x_{1} - P x_{2} + f (t),$ that is,

x' = P (t) x + f (t),

$x^{′} = P (t) x + f (t),$ (32)

where

x = [x 1 x 2] = [y y'], P (t) = [0 - Q 1 - P], and f (t) = [0 f (t)] .

$x = [\begin{array}{r} x_{1} \\ x_{2} \end{array}] = [\begin{array}{c} y \\ y^{′} \end{array}], P (t) = [\begin{array}{r} 0 & 1 \\ - Q & - P \end{array}], and f (t) = [\begin{array}{c} 0 \\ f (t) \end{array}] .$

Now two linearly independent solutions y1 $y_{1}$ and y2 $y_{2}$ of the homogeneous system y''+Py'+Qy=0 $y ″ + P y ′ + Q y = 0$ associated with (31) provide two linearly independent solutions

x 1 = [y 1 y' 1] and x 2 = [y 2 y' 2]

$x_{1} = [\begin{array}{r} y_{1} \\ y_{1}^{′} \end{array}] and x_{2} = [\begin{array}{r} y_{2} \\ y_{2}^{′} \end{array}]$

of the homogeneous system x'=P(t)x $x ′ = P (t) x$ associated with (32). Observe that the determinant of the fundamental matrix Φ=[x1x2] $Φ = [\begin{array}{r} x_{1} & x_{2} \end{array}]$ is simply the Wronskian

W = ∣ ∣ ∣ y 1 y' 1 y 2 y' 2 ∣ ∣ ∣

$W = | \begin{array}{l} y_{1} & y_{2} \\ y_{1}^{′} & y_{2}^{′} \end{array} |$

of the solutions y1 $y_{1}$ and y2 $y_{2}$ , so the inverse fundamental matrix is

Φ - 1 = 1 W ∣ ∣ ∣ y' 2 - y' 1 - y 2 y 1 ∣ ∣ ∣ .

$Φ^{- 1} = \frac{1}{W} | \begin{array}{r} y_{2}^{′} & - y_{2} \\ - y_{1}^{′} & y_{1} \end{array} | .$

Therefore the variation of parameters formula xp=Φ∫Φ−1f dt $x_{p} = Φ \int Φ^{- 1} f d t$ in (22) yields

[y p y' p] = = [y 1 y' 1 y 2 y' 2] \int 1 W [y' 2 - y' 1 - y 2 y 1] [0 f] d t [y 1 y' 1 y 2 y' 2] \int 1 W [- y 2 f y 1 f] d t .

$\begin{array}{rcl} [\begin{array}{r} y_{p} \\ y_{p}^{′} \end{array}] & = & [\begin{array}{r} y_{1} & y_{2} \\ y_{1}^{′} & y_{2}^{′} \end{array}] \int \frac{1}{W} [\begin{array}{r} y_{2}^{′} & - y_{2} \\ - y_{1}^{′} & y_{1} \end{array}] [\begin{array}{r} 0 \\ f \end{array}] d t \\ = & [\begin{array}{r} y_{1} & y_{2} \\ y_{1}^{′} & y_{2}^{′} \end{array}] \int \frac{1}{W} [\begin{array}{r} - y_{2} f \\ y_{1} f \end{array}] d t . \end{array}$

The first component of this column vector is

y p = [y 1 y 2] \int 1 W [- y 2 f y 1 f] d t = - y 1 \int y 2 f W d t + y 2 \int y 1 f W d t .

$y_{p} = [\begin{array}{r} y_{1} & y_{2} \end{array}] \int \frac{1}{W} [\begin{array}{r} - y_{2} f \\ y_{1} f \end{array}] d t = - y_{1} \int \frac{y_{2} f}{W} d t + y_{2} \int \frac{y_{1} f}{W} d t .$

If, finally, we supply the independent variable t throughout, the final result on the right-hand side here is simply the variation of parameters formula in Eq. (33) of Section 5.5 (where, however, the independent variable is denoted by x).

8.2 Problems

Apply the method of undetermined coefficients to find a particular solution of each of the systems in Problems 1 through 14. If initial conditions are given, find the particular solution that satisfies these conditions. Primes denote derivatives with respect to t.

x'=x+2y+3, y'=2x+y−2 $x ′ = x + 2 y + 3, y ′ = 2 x + y - 2$
x'=2x+3y+5, y'=2x+y−2t $x ′ = 2 x + 3 y + 5, y ′ = 2 x + y - 2 t$
x'=3x+4y, y'=3x+2y+t2;x(0)=y(0)=0 $x ′ = 3 x + 4 y, y ′ = 3 x + 2 y + t^{2}; x (0) = y (0) = 0$
x'=4x+y+et, y'=6x−y−et;x(0)=y(0)=1 $x ′ = 4 x + y + e^{t}, y ′ = 6 x - y - e^{t}; x (0) = y (0) = 1$
x'=6x−7y+10, y'=x−2y−2e−t $x ′ = 6 x - 7 y + 10, y ′ = x - 2 y - 2 e^{- t}$
x'=9x+y+2et, y'=−8x−2y+tet $x ′ = 9 x + y + 2 e^{t}, y ′ = - 8 x - 2 y + t e^{t}$
x'=−3x+4y+sin t, y'=6x−5y;x(0)=1, y(0)=0 $x ′ = - 3 x + 4 y + \sin t, y ′ = 6 x - 5 y; x (0) = 1, y (0) = 0$
x'=x−5y+2sin t, y'=x−y−3cos t $x ′ = x - 5 y + 2 \sin t, y ′ = x - y - 3 \cos t$
x'=x−5y+cos 2t, y'=x−y $x ′ = x - 5 y + \cos 2 t, y ′ = x - y$
x'=x−2y, y'=2x−y+et sin t $x ′ = x - 2 y, y ′ = 2 x - y + e^{t} \sin t$
x'=2x+4y+2, y'=x+2y+3;x(0)=1, y(0)=−1 $x ′ = 2 x + 4 y + 2, y ′ = x + 2 y + 3; x (0) = 1, y (0) = - 1$
x'=x+y+2t, y'=x+y−2t $x ′ = x + y + 2 t, y ′ = x + y - 2 t$
x'=2x+y+2et, y'=x+2y−3et $x ′ = 2 x + y + 2 e^{t}, y ′ = x + 2 y - 3 e^{t}$
x'=2x+y+1, y'=4x+2y+e4t $x ′ = 2 x + y + 1, y ′ = 4 x + 2 y + e^{4 t}$

Two Brine Tanks

Problems 15 and 16 are similar to Example 2, but with two brine tanks (having volumes V1 $V_{1}$ and V2 $V_{2}$ gallons as in Fig. 8.2.1) instead of three tanks. Each tank initially contains fresh water, and the inflow to tank 1 at the rate of r gallons per minute has a salt concentration of c0 $c_{0}$ pounds per gallon. (a) Find the amounts x1(t) $x_{1} (t)$ and x2(t) $x_{2} (t)$ of salt in the two tanks after t minutes. (b) Find the limiting (long-term) amount of salt in each tank. (c) Find how long it takes for each tank to reach a salt concentration of 1 lb/gal.

V1=100 $V_{1} = 100$ , V2=200 $V_{2} = 200$ , r=10 $r = 10$ , c0=2 $c_{0} = 2$
V1=200 $V_{1} = 200$ , V2=100 $V_{2} = 100$ , r=10 $r = 10$ , c0=3 $c_{0} = 3$

In Problems 17 through 34, use the method of variation of parameters (and perhaps a computer algebra system) to solve the initial value problem

x' = A x + f (t), x (a) = x a .

$x^{′} = A x + f (t), x (a) = x_{a} .$

In each problem we provide the matrix exponential eAt $e^{A t}$ as provided by a computer algebra system.

A=[61−7−2], f(t)=[6090], x(0)=[00],eAt=16[−e−t+7e5t−e−t+e5t7e−t−7e5t7e−t−e5t] $\begin{array}{l} A = [\begin{array}{r} 6 & - 7 \\ 1 & - 2 \end{array}], f (t) = [\begin{array}{r} 60 \\ 90 \end{array}], x (0) = [\begin{array}{r} 0 \\ 0 \end{array}], \\ e^{A t} = \frac{1}{6} [\begin{array}{c} - e^{- t} + 7 e^{5 t} & 7 e^{- t} - 7 e^{5 t} \\ - e^{- t} + e^{5 t} & 7 e^{- t} - e^{5 t} \end{array}] \end{array}$
Repeat Problem 17, but with f(t) replaced with [100t−50t]. $[\begin{array}{r} 100 t \\ - 50 t \end{array}] .$
A=[122−2], f(t)=[180t90], x(0)=[00],eAt=15[e−3t−2e−3t++4e2t2e2t−2e−3t+2e2t4e−3t+e2t] $\begin{array}{l} A = [\begin{array}{r} 1 & 2 \\ 2 & - 2 \end{array}], f (t) = [\begin{array}{r} 180 t \\ 90 \end{array}], x (0) = [\begin{array}{r} 0 \\ 0 \end{array}], \\ e^{A t} = \frac{1}{5} [\begin{array}{r} e^{- 3 t} & + & 4 e^{2 t} & - 2 e^{- 3 t} + 2 e^{2 t} \\ - 2 e^{- 3 t} & + & 2 e^{2 t} & 4 e^{- 3 t} + e^{2 t} \end{array}] \end{array}$
Repeat Problem 19, but with f(t) replaced with [75e2t0]. $[\begin{array}{c} 75 e^{2 t} \\ 0 \end{array}] .$
A=[45−1−2], f(t)=[18e2t30e2t], x(0)=[00],eAt=14[−e−t+5e3t−5e−t+5e5te−t−e3t5e−t−e3t] $\begin{array}{l} A = [\begin{array}{r} 4 & - 1 \\ 5 & - 2 \end{array}], f (t) = [\begin{array}{r} 18 e^{2 t} \\ 30 e^{2 t} \end{array}], x (0) = [\begin{array}{r} 0 \\ 0 \end{array}], \\ e^{A t} = \frac{1}{4} [\begin{array}{r} - e^{- t} + 5 e^{3 t} & e^{- t} - e^{3 t} \\ - 5 e^{- t} + 5 e^{5 t} & 5 e^{- t} - e^{3 t} \end{array}] \end{array}$
Repeat Problem 21, but with f(t) replaced with [28e−t20e3t]. $[\begin{array}{r} 28 e^{- t} \\ 20 e^{3 t} \end{array}] .$
A=[39−1−3],f(t)=[75], x(0)=[35],eAt=[1+3t9t−t1−3t] $\begin{array}{l} A = [\begin{array}{r} 3 & - 1 \\ 9 & - 3 \end{array}], f (t) = [\begin{array}{r} 7 \\ 5 \end{array}], x (0) = [\begin{array}{r} 3 \\ 5 \end{array}], \\ e^{A t} = [\begin{array}{c} 1 + 3 t & - t \\ 9 t & 1 - 3 t \end{array}] \end{array}$
Repeat Problem 23, but with f(t)=[0t−2] $f (t) = [\begin{array}{c} 0 \\ t^{- 2} \end{array}]$ and x(1)=[37] $x (1) = [\begin{array}{r} 3 \\ 7 \end{array}]$ .
A=[21−5−2],f(t)=[4t1], x(0)=[00],eAt=[cos t+2 sin tsin t−5 sin tcos−2 sin t] $\begin{array}{l} A = [\begin{array}{r} 2 & - 5 \\ 1 & - 2 \end{array}], f (t) = [\begin{array}{r} 4 t \\ 1 \end{array}], x (0) = [\begin{array}{r} 0 \\ 0 \end{array}], \\ e^{A t} = [\begin{array}{c} \cos t + 2 \sin t & - 5 \sin t \\ \sin t & \cos - 2 \sin t \end{array}] \end{array}$
Repeat Problem 25, but with f(t)=[4 cos t6 sin t] $f (t) = [\begin{array}{r} 4 \cos t \\ 6 \sin t \end{array}]$ and x(0)=[35] $x (0) = [\begin{array}{r} 3 \\ 5 \end{array}]$ .
A=[21−4−2], f(t)=[36t26t], x(0)=[00],eAt=[1+2tt−4t1−2t] $\begin{array}{l} A = [\begin{array}{r} 2 & - 4 \\ 1 & - 2 \end{array}], f (t) = [\begin{array}{c} 36 t^{2} \\ 6 t \end{array}], x (0) = [\begin{array}{r} 0 \\ 0 \end{array}], \\ e^{A t} = [\begin{array}{c} 1 + 2 t & - 4 t \\ t & 1 - 2 t \end{array}] \end{array}$
Repeat Problem 27, but with f(t)=[4ln tt−1] $f (t) = [\begin{array}{c} 4 \ln t \\ t^{- 1} \end{array}]$ and x(1)=[1−1]. $x (1) = [\begin{array}{r} 1 \\ - 1 \end{array}] .$
A=[01−10], f(t)=[sec t0], x(0)=[00],eAt=[cos tsin t−sin tcos t] $\begin{array}{l} A = [\begin{array}{r} 0 & - 1 \\ 1 & 0 \end{array}], f (t) = [\begin{array}{c} \sec t \\ 0 \end{array}], x (0) = [\begin{array}{r} 0 \\ 0 \end{array}], \\ e^{A t} = [\begin{array}{r} \cos t & - \sin t \\ \sin t & \cos t \end{array}] \end{array}$
A=[02−20], f(t)=[t cos 2tt sin 2t], x(0)=[00],eAt=[cos 2tsin 2t−sin 2tcos 2t] $\begin{array}{l} A = [\begin{array}{r} 0 & - 2 \\ 2 & 0 \end{array}], f (t) = [\begin{array}{r} t \cos 2 t \\ t \sin 2 t \end{array}], x (0) = [\begin{array}{r} 0 \\ 0 \end{array}], \\ e^{A t} = [\begin{array}{r} \cos 2 t & - \sin 2 t \\ \sin 2 t & \cos 2 t \end{array}] \end{array}$
A=⎡⎣⎢100210321⎤⎦⎥, f(t)=⎡⎣⎢006et⎤⎦⎥, x(0)=⎡⎣⎢000⎤⎦⎥,eAt=⎡⎣⎢et002tetet0(3t+2t2)et2tetet⎤⎦⎥ $\begin{array}{l} A = [\begin{array}{r} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}], f (t) = [\begin{array}{c} 0 \\ 0 \\ 6 e^{t} \end{array}], x (0) = [\begin{array}{r} 0 \\ 0 \\ 0 \end{array}], \\ e^{A t} = [\begin{array}{c} e^{t} & 2 t e^{t} & (3 t + 2 t^{2}) e^{t} \\ 0 & e^{t} & 2 t e^{t} \\ 0 & 0 & e^{t} \end{array}] \end{array}$
A=⎡⎣⎢100310432⎤⎦⎥, f(t)=⎡⎣⎢002e2t⎤⎦⎥, x(0)=⎡⎣⎢000⎤⎦⎥,eAt=⎡⎣⎢et003tetet0(−13−9t)et+13e2t−3et+3e2te2t⎤⎦⎥ $\begin{array}{l} A = [\begin{array}{r} 1 & 3 & 4 \\ 0 & 1 & 3 \\ 0 & 0 & 2 \end{array}], f (t) = [\begin{array}{c} 0 \\ 0 \\ 2 e^{2 t} \end{array}], x (0) = [\begin{array}{r} 0 \\ 0 \\ 0 \end{array}], \\ e^{A t} = [\begin{array}{c} e^{t} & 3 t e^{t} & (- 13 - 9 t) e^{t} + 13 e^{2 t} \\ 0 & e^{t} & - 3 e^{t} + 3 e^{2 t} \\ 0 & 0 & e^{2 t} \end{array}] \end{array}$
A=⎡⎣⎢⎢⎢⎢0000400033000840⎤⎦⎥⎥⎥⎥, f(t)=30 ⎡⎣⎢⎢⎢⎢tttt⎤⎦⎥⎥⎥⎥, x(0)=⎡⎣⎢⎢⎢⎢0000⎤⎦⎥⎥⎥⎥,eAt=⎡⎣⎢⎢⎢⎢10004t1008t+6t23t1032t2+8t28t+6t24t1⎤⎦⎥⎥⎥⎥ $\begin{array}{l} A = [\begin{array}{r} 0 & 4 & 3 & 0 \\ 0 & 0 & 3 & 8 \\ 0 & 0 & 0 & 4 \\ 0 & 0 & 0 & 0 \end{array}], f (t) = 30 [\begin{array}{r} t \\ t \\ t \\ t \end{array}], x (0) = [\begin{array}{r} 0 \\ 0 \\ 0 \\ 0 \end{array}], \\ e^{A t} = [\begin{array}{c} 1 & 4 t & 8 t + 6 t^{2} & 32 t^{2} + 8 t^{2} \\ 0 & 1 & 3 t & 8 t + 6 t^{2} \\ 0 & 0 & 1 & 4 t \\ 0 & 0 & 0 & 1 \end{array}] \end{array}$
A=⎡⎣⎢⎢⎢⎢0000400080200842⎤⎦⎥⎥⎥⎥, f(t)=⎡⎣⎢⎢⎢⎢06t0e2t⎤⎦⎥⎥⎥⎥, x(0)=⎡⎣⎢⎢⎢⎢4221⎤⎦⎥⎥⎥⎥,eAt=⎡⎣⎢⎢⎢⎢10004t1004(−1+e2t)0e2t016t(−1+e2t)4(−1+e2t)4te2te2t⎤⎦⎥⎥⎥⎥ $\begin{array}{l} A = [\begin{array}{r} 0 & 4 & 8 & 0 \\ 0 & 0 & 0 & 8 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 2 \end{array}], f (t) = [\begin{array}{r} 0 \\ 6 t \\ 0 \\ e^{2 t} \end{array}], x (0) = [\begin{array}{r} 4 \\ 2 \\ 2 \\ 1 \end{array}], \\ e^{A t} = [\begin{array}{c} 1 & 4 t & 4 (- 1 + e^{2 t}) & 16 t (- 1 + e^{2 t}) \\ 0 & 1 & 0 & 4 (- 1 + e^{2 t}) \\ 0 & 0 & e^{2 t} & 4 t e^{2 t} \\ 0 & 0 & 0 & e^{2 t} \end{array}] \end{array}$

8.2 Application Automated Variation of Parameters

The application of the variation of parameters formula in Eq. (28) encourages so mechanical an approach as to encourage especially the use of a computer algebra system. The following Mathematica commands were used to check the results in Example 4 of this section.


A = {{4,2}, {3,−1}};
x0 = {{7}, {3}};
f[t_] := {{−15 t Exp[−2t]},{−4 t Exp[−2t]}};
exp[A_] := MatrixExp[A]
x = exp[A∗t].(x0 + Integrate[exp[−A∗s].f[s], {s,0,t}])

The matrix exponential commands illustrated in the Section 5.6 application provide the basis for analogous Maple and Matlab computations. You can then check routinely the answers for Problems 17 through 34 of this section.

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Table of Contents for 8.2 Nonhomogeneous Linear Systems

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Table of Contents for
8.2 Nonhomogeneous Linear Systems