10. 4 Nonhomogeneous Linear Systems

INTRODUCTION

The methods of undetermined coefficients and variation of parameters used in Chapter 3 to find particular solutions of nonhomogeneous linear ODEs can both be adapted to the solution of nonhomogeneous linear systems. Of the two methods, variation of parameters is the more powerful technique. However, there are instances when the method of undetermined coefficients provides a quick means of finding a particular solution.

In Section 10.1, we saw that the general solution of a nonhomogeneous linear system X′ = AX + F(t) on an interval I is X = Xc + Xp where Xc = c1X1 + c2X2 + + cnXn is the complementary function or general solution of the associated homogeneous linear system X′ = AX and Xp is any particular solution of the nonhomogeneous system. We just saw how to obtain Xc in Section 10.2 when A was an n × n matrix of constants; we now consider three methods for obtaining Xp.

10.4.1 Undetermined Coefficients

The Assumptions

As in Section 3.4, the method of undetermined coefficients consists of making an educated guess about the form of a particular solution vector Xp; the guess is motivated by the types of functions that comprise the entries of the column matrix F(t). Not surprisingly, the matrix version of undetermined coefficients is only applicable to X′ = AX + F(t) when the entries of A are constants and the entries of F(t) are constants, polynomials, exponential functions, sines and cosines, or finite sums and products of these functions.

EXAMPLE 1 Undetermined Coefficients

Solve the system X′ = X + on the interval (−∞, ∞).

SOLUTION

We first solve the associated homogeneous system

The characteristic equation of the coefficient matrix A,

yields the complex eigenvalues λ1 = i and λ2 = = −i. By the procedures of Section 10.2, we find

Now since F(t) is a constant vector, we assume a constant particular solution vector Xp = . Substituting this latter assumption into the original system and equating entries leads to

Solving this algebraic system gives a1 = 14 and b1 = 11, and so a particular solution is Xp = . The general solution of the original system of differential equations on the interval (−∞, ∞) is then X = Xc + Xp or

≡

EXAMPLE 2 Undetermined Coefficients

Solve the system on the interval (−∞, ∞).

SOLUTION

The eigenvalues and corresponding eigenvectors of the associated homogeneous system X′ = X are found to be λ1 = 2, λ2 = 7, K1 = , and K2 = . Hence the complementary function is

Now because F(t) can be written F(t) = t + we shall try to find a particular solution of the system that possesses the same form:

Substituting this last assumption into the given system yields

or

From the last identity we obtain four algebraic equations in four unknowns:

Solving the first two equations simultaneously yields a2 = −2 and b2 = 6. We then substitute these values into the last two equations and solve for a1 and b1. The results are a1 = −, b1 = . It follows, therefore, that a particular solution vector is

The general solution of the system on the interval (−∞, ∞) is X = Xc + Xp or

≡

EXAMPLE 3 Form of Xp

Determine the form of a particular solution vector Xp for the system

SOLUTION

Because F(t) can be written in matrix terms as

a natural assumption for a particular solution would be

≡

REMARKS

The method of undetermined coefficients for linear systems is not as straightforward as the last three examples would seem to indicate. In Section 3.4, the form of a particular solution yp was predicated on prior knowledge of the complementary function yc. The same is true for the formation of Xp. But there are further difficulties; the special rules governing the form of yp in Section 3.4 do not quite carry to the formation of Xp. For example, if F(t) is a constant vector as in Example 1 and λ = 0 is an eigenvalue of multiplicity one, then Xc contains a constant vector. Under the “multiplication rule” on page 137, we would ordinarily try a particular solution of the form Xp = t. This is not the proper assumption for linear systems; it should be Xp = . Similarly, in Example 3, if we replace e–t in F(t) by e2t (λ = 2 is an eigenvalue), then the correct form of the particular solution vector is

Rather than delve into these difficulties, we turn instead to the method of variation of parameters.

10.4.2 Variation of Parameters

A Fundamental Matrix

If X1, X2, … , Xn is a fundamental set of solutions of the homogeneous system X′ = AX on an interval I, then its general solution on the interval is the linear combination X = c1X1 + c2X2 + + cnXn , or

(1)

The last matrix in (1) is recognized as the product of an n × n matrix with an n × 1 matrix. In other words, the general solution (1) can be written as the product

X = Φ(t)C,(2)

where C is the n × 1 column vector of arbitrary constants c1, c2, … , cn, and Φ(t) is the n × n matrix whose columns consist of the entries of the solution vectors of the system X′ = AX:

Φ(t) = .

The matrix Φ(t) is called a fundamental matrix of the system on the interval.

In the discussion that follows, we need to use two properties of a fundamental matrix:

  • A fundamental matrix Φ(t) is nonsingular.
  • If Φ(t) is a fundamental matrix of the system X′ = AX, then

Φ′(t) = AΦ(t).(3)

A reexamination of (12) of Theorem 10.1.3 shows that det Φ(t) is the same as the Wronskian W(X1, X2, … , Xn). Hence the linear independence of the columns of Φ(t) on the interval I guarantees that det Φ(t) ≠ 0 for every t in the interval. Since Φ(t) is nonsingular, the multiplicative inverse Φ−1(t) exists for every t in the interval. The result given in (3) follows immediately from the fact that every column of Φ(t) is a solution vector of X′ = AX.

Variation of Parameters

Analogous to the procedure in Section 3.5, we ask whether it is possible to replace the matrix of constants C in (2) by a column matrix of functions

so that Xp = Φ(t)U(t)(4)

is a particular solution of the nonhomogeneous system

X′ = AX + F(t).(5)

By the Product Rule, the derivative of the last expression in (4) is

X′p = Φ(t)U′(t) + Φ′(t)U(t).(6)

Note that the order of the products in (6) is very important. Since U(t) is a column matrix, the products U′(t)Φ(t) and U(t)Φ′(t) are not defined. Substituting (4) and (6) into (5) gives

Φ(t)U′(t) + Φ′(t)U(t) = AΦ(t)U(t) + F(t).(7)

Now if we use (3) to replace Φ′(t), (7) becomes

Φ(t)U′(t) + AΦ(t)U(t) = AΦ(t)U(t) + F(t)

or Φ(t)U′(t) = F(t).(8)

Multiplying both sides of equation (8) by Φ−1(t) gives

U′(t) = Φ−1(t)F(t) and so U(t) = Φ−1(t)F(t) dt.

Since Xp = Φ(t)U(t), we conclude that a particular solution of (5) is

Xp = Φ(t) Φ–1(t)F(t) dt.(9)

To calculate the indefinite integral of the column matrix Φ−1(t)F(t) in (9), we integrate each entry. Thus the general solution of the nonhomogeneous system (5) is X = Xc + Xp or

X = Φ(t)C + Φ(t) Φ–1(t)F(t) dt.(10)

EXAMPLE 4 Variation of Parameters

Find the general solution of the nonhomogeneous system

(11)

on the interval (−∞, ∞).

SOLUTION

We first solve the associated homogeneous system

(12)

The characteristic equation of the coefficient matrix is

det(A − λI) = = (λ + 2)(λ + 5) = 0,

so the eigenvalues are λ1 = −2 and λ2 = −5. By the usual method we find that the eigenvectors corresponding to λ1 and λ2 are, respectively,

and

The solution vectors of the homogeneous system (12) are then

The entries in X1 form the first column of Φ(t), and the entries in X2 form the second column of Φ(t). Hence

Φ(t) = and Φ−1(t) =

From (9) we obtain

Hence from (10), the general solution of (11) on the interval (−∞, ∞) is

≡

Initial-Value Problem

The general solution of the nonhomogeneous system (5) on an interval can be written in an alternative manner

(13)

where t and t0 are points in the interval I. The last form is useful in solving (5) subject to an initial condition X(t0) = X0, because the limits of integration are chosen so that the particular solution vanishes at t = t0. Substituting t = t0 in (13) yields X0 = Φ (t0)C, from which we get C = Φ–1(t0)X0. Substituting this last result in (13) gives the following solution of the initial-value problem:

(14)

10.4.3 Diagonalization

The Assumptions

As in Section 10.3, if the coefficient matrix A possesses n linearly independent eigenvectors, then we can use diagonalization to uncouple the system X′ = AX + F(t). Suppose P is the matrix such that P−1AP = D, where D is a diagonal matrix. Substituting X = PY into the nonhomogeneous system X′ = AX + F(t) gives

PY′ = APY + F or Y′ = P−1APY + P−1F or Y′ = DY + G.(15)

In the last equation in (15), G = P–1F is a column vector. So each differential equation in this new system has the form = λi yi + gi(t), i = 1, 2, … , n. But notice that, unlike the procedure for solving a homogeneous system X′ = AX, we now are required to compute the inverse of the matrix P.

EXAMPLE 5 Diagonalization

Solve by diagonalization.

SOLUTION

The eigenvalues and corresponding eigenvectors of the coefficient matrix are found to be λ1 = 0, λ2 = 5, K1 = , K2 = . Thus we find P = and P–1 = . Using the substitution X = PY and

the uncoupled system is

The solutions of the two linear first-order differential equations

= et and = 5y2 + et

are, respectively, y1 = et + c1 and y2 = −et + c2e5t. Hence, the solution of the original system is

(16)

Written in the usual manner using column vectors, (16) is

≡

10.4 Exercises Answers to selected odd-numbered problems begin on page ANS-28.

10.4.1 Undetermined Coefficients

In Problems 1–8, use the method of undetermined coefficients to solve the given system.



In Problems 9 and 10, solve the given initial-value problem.

  3. Consider the large mixing tanks shown in FIGURE 10.4.1. Suppose that both tanks A and B initially contain 100 gallons of brine. Liquid is pumped in and out of the tanks as indicated in the figure; the mixture pumped between and out of the tanks is assumed to be well-stirred.
    1. Construct a mathematical model in the form of a linear system of first-order differential equations for the number of pounds x1(t) and x2(t) of salt in tanks A and B, respectively, at time t. Write the system in matrix form. [Hint: See Section 2.9 and Problem 52, Chapter 2 in Review.]
    2. Use the method of undetermined coefficients to solve the linear system in part (a) subject to x1(0) = 60, x2(0) = 10.
    3. What are ? Interpret this result.
    4. Use a graphing utility or CAS to plot the graphs of x1(t) and x2(t) in the same coordinate plane.
      Two tanks are labeled A and B. 6 pipes are connected to the tanks and the rate of flow is as follows. Into A: pure water at 2 gallons per minute. From A to B: mixture at 2 gallons per minute. Out from B: mixture at 3 gallons per minute. Into B: half a pound per gallon at 2 gallons per minute. From B to A: mixture at 1 gallon per minute. Out from A: mixture at 1 gallon per minute.

      FIGURE 10.4.1 Mixing tanks in Problem 11

    1. The system of differential equations for the currents i2(t) and i3(t) in the electrical network shown in FIGURE 10.4.2 is

      Use the method of undetermined coefficients to solve the system if R1 = 2 Ω, R2 = 3 Ω, L1 = 1 h, L2 = 1 h, E = 60 V, i2(0) = 0, and i3(0) = 0.

    2. Determine the current i1(t).
      A circuit diagram with two resistors, R subscript 1 and R subscript 2 and two inductors, L subscript 1 and L subscript 2. R subscript 1 is in series with 2 parallel paths. The first one consists of L subscript 1. The second path consists of R subscript 2 and L subscript 2 in series. L subscript 1 and L subscript are in parallel. Current i subscript 1 from source E flows through R subscript 1 and branches into i subscript 2 and i subscript 3. i subscript 2 flows through the path with L subscript 1. i subscript 3 flows through the path with R subscript 2 and L subscript 2.

      FIGURE 10.4.2 Network in Problem 12

10.4.2 Variation of Parameters

In Problems 13–32, use variation of parameters to solve the given system.



In Problems 33 and 34, use (14) to solve the given initial-value problem.

  3. The system of differential equations for the currents i1(t) and i2(t) in the electrical network shown in FIGURE 10.4.3 is

    Use variation of parameters to solve the system if R1 = 8 Ω, R2 = 3 Ω, L1 = 1 h, L2 = 1 h, E(t) = 100 sin t V, i1(0) = 0, and i2(0) = 0.

    A circuit diagram with two resistors, R subscript 1 and R subscript 2, and two inductors, L subscript 1 and L subscript 2. R subscript 1 is in series with 2 parallel paths. The first one consists of L subscript 1 and L subscript 2 in series. The second path consists of R subscript 2 and L subscript 2 in series; R subscript 2 is in parallel to L subscript 1. Current i subscript 1 from source E flows through R subscript 1 and branches into i subscript 2 and i subscript 3. i subscript 2 flows through the path with L subscript 1 and L subscript 2. i subscript 3 flows through the path with R subscript 2 and L subscript 2.

    FIGURE 10.4.3 Network in Problem 35

Computer Lab Assignment

  1. Solving a nonhomogeneous linear system X′ = AX + F(t) by variation of parameters when A is a 3 × 3 (or larger) matrix is almost an impossible task to do by hand. Consider the system
    1. Use a CAS or linear algebra software to find the eigenvalues and eigenvectors of the coefficient matrix.
    2. Form a fundamental matrix Φ(t) and use the computer to find Φ−1(t).
    3. Use the computer to carry out the computations of Φ–1(t)F(t), Φ–1(t)F(t) dt, Φ(t) Φ–1(t)F(t) dt, Φ(t)C, and Φ(t)C + Φ–1(t)F(t) dt, where C is a column matrix of constants c1, c2, c3, and c4.
    4. Rewrite the computer output for the general solution of the system in the form X = Xc + Xp, where Xc = c1X1 + c2X2 + c3X3 + c4X4.

10.4.3 Diagonalization

In Problems 37–40, use diagonalization to solve the given system.