The Method of Undetermined Coefficients

Theorem 3.2.7 Abel’s Theorem) 4

4.3 The Method of Undetermined Coefficients

d. Now suppose that the initial conditions are

u₁( 0)= −2, u^′₁( 0)=0, u₂( 0) =1, u^′₂( 0)=0. (25) Proceed as in partcto show that the corresponding solutions are u₁(t) = −2 cos

6t

andu₂(t) =cos 6t

.

e. Observe that the solutions obtained in partscandddescribe two distinct modes of vibration. In the first, the frequency of the motion is 1, and the two masses move in phase, both moving up or down together; the second mass moves twice as far as the first. The second motion has frequency

6, and the masses move out of phase with each other, one moving down while the other is moving up, and vice versa. In this mode the first mass moves twice as far as the second. For other initial conditions, not proportional to either of equations (24) or (25), the motion of the masses is a combination of these two modes.

u1

k1 = 3

m1 = 1

m2 = 1 k2 = 2

u2

FIGURE 4.2.4 A two-spring, two-mass system.

30. In this problem we outline one way to show that ifr₁, . . .,rn

are all real and different, thene^r1t, . . .,e^rntare linearly independent on−∞< t< ∞. To do this, we consider the linear relation

c₁e^r1t+ · · · +cne^rnt=0, −∞< t<∞ (26) and show that all the constants are zero.

a. Multiply equation (26) bye^−r1tand differentiate with respect tot, thereby obtaining

c₂(r₂−r₁)e^(r2−r1)t+ · · · +cn(rn−r₁)e^(rn−r1)t=0.

b. Multiply the result of partabye^{−(r2−r1)t} and differentiate with respect totto obtain

c₃(r₃−r₂) (r₃−r₁)e^(r3−r2)t

+ · · · +c_n(r_n−r₂) (r_n−r₁)e^(rn−r2)t=0.

c. Continue the procedure from parts a and b, eventually obtaining

c_n(r_n−r_n−1)· · ·(r_n−r₁)e^{(rn−rn−1)t}=0.

Hencec_n=0, and therefore,

c₁e^r1t+ · · · +c_n−1e^rn−1t=0.

d. Repeat the preceding argument to show thatc_n−1 = 0. In a similar way it follows thatc_n−2 = · · · = c₁ = 0. Thus the functionse^r1t,. . .,e^rntare linearly independent.

31. In this problem we indicate one way to show that ifr = r₁is a root of multiplicitys of the characteristic polynomialZ(r) , then e^r1t, t e^r1t, . . ., t^s−1e^r1tare solutions of equation (1). This problem extends ton^thorder equations the method for second-order equations given in Problem17of Section 3.4. We start from equation (2) in the text

L[e^{r t}]=e^{r t}Z(r) (27) and differentiate repeatedly with respect tor, settingr =r₁after each differentiation.

a. Recall that ifr₁is a root of multiplicitys, then

Z(r) = (r−r₁)^sq(r) , whereq(r) is a polynomial of degree n−sandq(r₁) =0. Show thatZ(r₁) ,Z^′(r₁) , . . .,Z^(s−1)(r₁) are all zero, butZ^(s)(r₁) =0.

b. By differentiating equation (27) repeatedly with respect tor, show that

∂

∂rL[e^{r t}]=L ∂

∂re^{r t}

=L[t e^{r t}], ..

.

∂^s−1

∂r^s−1L[e^{r t}]=L[t^s−1e^{r t}].

c. Show that e^r1t,t e^r1t, . . .,t^s−1e^r1t are solutions of equation (27).

Just as for the second-order linear differential equation, when the constant coefficient linear differential operator L is applied to a polynomial A0t^m + A1t^m−1 + · · · + Am, an exponential function e^αt, a linear combination of sine and cosine functionsa1cos(βt) + a2sin(βt) , the result is a polynomial, an exponential function, or a linear combination of sine and cosine functions, respectively. Hence, ifg(t) is a sum of polynomials, exponentials, sines, and cosines, or products of such functions, we can expect that it is possible to findY(t) by choosing a suitable combination of polynomials, exponentials, and so forth, multiplied by a number of undetermined constants. The constants are then determined by substituting the assumed expression into the nonhomogeneous linear differential equation (1).

The main difference in using this method for higher-order equations stems from the fact that roots of the characteristic polynomial equation may have multiplicity greater than 2. Consequently, terms proposed for the nonhomogeneous part of the solution may need to be multiplied by higher powers of t to make them different from terms in the solution of the corresponding homogeneous equation. The following examples illustrate this. In these examples we have omitted numerous straightforward algebraic steps, because our main goal is to show how to arrive at the correct form for the assumed solution.

EXAMPLE 1

Find the general solution of

y^′′′−3y^′′+3y^′−y=4e^t. (2)

Solution:

The characteristic polynomial for the homogeneous equation corresponding to equation (2) is r³−3r²+3r−1=(r−1)³,

so the general solution of the homogeneous equation is

yc(t)=c₁e^t+c₂t e^t+c₃t²e^t. (3) To find a particular solution Y(t) of equation (2), we start by assuming that Y(t) = Ae^t. However, sincee^t,t e^t, andt²e^tare all solutions of the homogeneous equation, we must multiply this initial choice byt³. Thus our final assumption is thatY(t) = At³e^t, where Ais an undetermined coefficient.

To find the correct value forA, differentiateY(t) three times, substitute foryand its derivatives in equation (2), and collect terms in the resulting equation. In this way we obtain

6Ae^t=4e^t. ThusA= 2

3and the particular solution is

Y(t) = 2

3t³e^t. (4)

The general solution of the nonhomogeneous differential equation (2) is the sum of yc(t) from equation (3) andY(t) from equation (4):

y=c₁e^t+c₂t e^t+c₃t²e^t+2 3t³e^t.

EXAMPLE 2

Find a particular solution of the equation

y^{( 4)}+2y^′′+y=3 sint−5 cost. (5)

▼

▼ _Solution:

The general solution of the homogeneous equation was found in Example 3 of Section 4.2; it is y_c(t) =c₁cost+c₂sint+c₃tcost+c₄tsint, (6) corresponding to the rootsr=i,i,−i, and−iof the characteristic equation. Our initial assumption for a particular solution isY(t) =Asint+Bcost, but we must multiply this choice byt²to make it different from all solutions of the homogeneous equation. Thus our final assumption is

Y(t) = At²sint+Bt²cost.

Next, we differentiateY(t) four times, substitute into the differential equation (5), and collect terms, obtaining finally

−8Asint−8Bcost=3 sint−5 cost.

ThusA= −3 8,B= 5

8, and the particular solution of equation (4) is Y(t)= −3

8t²sint+5

8t²cost. (7)

If g(t) is a sum of several terms, it may be easier in practice to compute separately the particular solution corresponding to each term ing(t) . In the same way as for second- order differential equations, the particular solution of the complete problem is the sum of the particular solutions of the individual component problems. This is illustrated in the following example.

EXAMPLE 3

Find a particular solution of

y^′′′−4y^′=t+3 cost+e^−2t. (8)

Solution:

First we solve the homogeneous equation. The characteristic equation isr³−4r =0, and the roots arer=0,±2; hence

y_c(t) =c₁+c₂e^2t+c₃e^−2t.

We can write a particular solution of equation (8) as the sum of particular solutions of the differential equations

y^′′′−4y^′=t, y^′′′−4y^′=3 cost, y^′′′−4y^′=e^−2t.

Our initial choice for a particular solutionY₁(t) of the first equation isA₀t+A₁, but a constant is a solution of the homogeneous equation, so we multiply byt. Thus

Y₁(t) =t(A₀t+A₁). For the second equation we choose

Y₂(t) =Bcost+Csint,

and there is no need to modify this initial choice since sint and cost are not solutions of the homogeneous equation. Finally, for the third equation, sincee^−2tis a solution of the homogeneous equation, we assume that

Y₃(t) =E t e^−2t.

The constants are determined by substituting into the individual differential equations; they are A₀= −1

8, A₁=0, B=0, C= −3

5, andE= 1

8. Hence a particular solution of equation (8) is Y(t) = −1

8t²− 3

5sint+ 1

8t e^−2t. (9)

You should keep in mind that the amount of algebra required to calculate the coefficients may be quite substantial for higher-order equations, especially if the nonhomogeneous term is even moderately complicated. A computer algebra system can be extremely helpful in executing these algebraic calculations.

The method of undetermined coefficients can be used whenever it is possible to guess the correct form forY(t) . However, this is usually impossible for differential equations not having constant coefficients, or for nonhomogeneous terms other than the type described previously.

For more complicated problems we can use the method of variation of parameters, which is discussed in the next section.

Problems

In each of Problems1through6, determine the general solution of the given differential equation.

1. y^′′′−y^′′−y^′+y=2e^−t+3 2. y^{( 4)} −y=3t+cost 3. y^′′′+y^′′+y^′+y=e^−t+4t 4. y^{( 4)} −4y^′′=t²+e^t 5. y^{( 4)} +2y^′′+y=3+cos 2t 6. y^{( 6)} +y^′′′=t

In each of Problems7through9, find the solution of the given initial- value problem. Then plot a graph of the solution.

G 7. y^′′′+4y^′=t; y( 0) =y^′( 0)=0, y^′′( 0)=1

G 8. y^{( 4)} +2y^′′+y=3t+4; y( 0) =y^′( 0)=0, y^′′( 0)=y^′′′( 0) =1

G 9. y^{( 4)} +2y^′′′+y^′′+8y^′−12y=12 sint−e^−t; y( 0)=3, y^′( 0) =0, y^′′( 0) = −1, y^′′′( 0)=2

In each of Problems 10through 13, determine a suitable form for Y(t) if the method of undetermined coefficients is to be used. Do not evaluate the constants.

10. y^′′′−2y^′′+y^′=t³+2e^t 11. y^′′′−y^′=t e^−t+2 cost

12. y^{( 4)} −y^′′′−y^′′+y^′=t²+4+tsint 13. y^{( 4)} +2y^′′′+2y^′′=3e^t+2t e^−t+e^−tsint

14. Consider the nonhomogeneous n^th order linear differential equation

a₀y⁽ⁿ⁾+a₁y⁽ⁿ⁻¹⁾+ · · · +a_ny=g(t) , (10) wherea₀,. . .,anare constants. Verify that ifg(t) is of the form

e^αt(b₀t^m+ · · · +b_m) ,

then the substitutiony=e^αtu(t) reduces equation (10) to the form k₀u⁽ⁿ⁾+k₁u⁽ⁿ⁻¹⁾+ · · · +k_nu=b₀t^m+ · · · +b_m, (11) wherek₀, . . .,k_nare constants. Determinek₀andk_nin terms of the a's andα. Thus the problem of determining a particular solution of the original equation is reduced to the simpler problem of determining a particular solution of an equation with constant coefficients and a polynomial for the nonhomogeneous term.

Method of Annihilators.In Problems15 through17, we consider another way of arriving at the proper form ofY(t) for use in the method of undetermined coefficients. The procedure is based on the

observation that exponential, polynomial, or sinusoidal terms (or sums and products of such terms) can be viewed as solutions of certain linear homogeneous differential equations with constant coefficients. It is convenient to use the symbolDfor d

dt. Then, for example,e^−t is a solution of (D+1)y= 0; the differential operatorD+1 is said to annihilate, or to be anannihilatorof,e^−t. In the same way,D²+4 is an annihilator of sin 2tor cos 2t, (D−3)² = D²−6D+9 is an annihilator ofe^3tort e^3t, and so forth.

15. Show that linear differential operators with constant coefficients obey the commutative law. That is, show that

(D−a) (D−b)f =(D−b) (D−a)f

for any twice-differentiable function fand any constantsaandb. The result extends at once to any finite number of factors.

16. Consider the problem of finding the form of a particular solution Y(t) of

(D−2)³(D+1)Y =3e^2t−t e^−t, (12) where the left-hand side of the equation is written in a form corresponding to the factorization of the characteristic polynomial.

a. Show thatD−2 and (D+1)², respectively, are annihilators of the terms on the right-hand side of equation (12), and that the combined operator (D−2) (D+1)²annihilates both terms on the right-hand side of equation (12) simultaneously.

b. Apply the operator (D−2) (D+1)²to equation (12) and use the result of Problem15to obtain

(D−2)⁴(D+1)³Y =0. (13) Thus Y is a solution of the homogeneous equation (13). By solving equation (13), show that

Y(t) =c₁e^2t+c₂t e^2t+c₃t²e^2t+c₄t³e^2t+c₅e^−t +c₆t e^−t+c₇t²e^−t, (14) wherec₁, . . .,c₇are constants, as yet undetermined.

c. Observe thate^2t, t e^2t, t²e^2t, and e^−t are solutions of the homogeneous equation corresponding to equation (12); hence these terms are not useful in solving the nonhomogeneous equation. Therefore, choose c₁, c₂, c₃, and c₅ to be zero in equation (14), so that

Y(t) =c₄t³e^2t+c₆t e^−t+c₇t²e^−t. (15) This is the form of the particular solutionY of equation (12).

The values of the coefficients c₄, c₆, andc₇ can be found by substituting from equation (15) in the differential equation (12).

Summary of the Method of Annihilators. Suppose that

L(D)y=g(t) , (16)

whereL(D) is a linear differential operator with constant coefficients, andg(t) is a sum or product of exponential, polynomial, or sinusoidal terms. To find the form of a particular solution of equation (16), you can proceed as follows:

a. Find a differential operator H(D) with constant coeffi- cients that annihilates g(t) ---that is, an operator such that H(D)g(t) =0.

b. ApplyH(D) to equation (16), obtaining

H(D)L(D)y=0, (17)

which is a homogeneous equation of higher-order.

c. Solve equation (17).

d. Eliminate from the solution found in stepcthe terms that also appear in the solution of L(D)y = 0. The remaining terms constitute the correct form of a particular solution of equation (16).

17. Use the method of annihilators to find the form of a particular solutionY(t) for each of the equations in Problems10through13. Do not evaluate the coefficients.