# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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EXAMPLE

2

EXAMPLE

3

where A is an undetermined coefficient. To find the correct value for A, we differentiate

Y(t) three times, substitute for y and its derivatives in Eq. (2), and collect terms in the resulting equation. In this way we obtain

6 Ael = 4e'.

Thus A = 3 and the particular solution is

Y(t) = 2 tV. (4)

The general solution of Eq. (2) is the sum of yc (t) from Eq. (3) and Y(t) from Eq. (4).

Find a particular solution of the equation

yiv + 2y" + y = 3sin t — 5cos t. (5)

The general solution of the homogeneous equation was found in Example 3 of Section 4.2, namely,

yc(t) = cx cos t + c2 sin t + c3tcos t + c4tsin t, (6)

corresponding to the roots r = i, i, —i, and —i of the characteristic equation. Our initial assumption for a particular solution is Y( t) = A sin t + B cos t, but we must multiply this choice by t2 to make it different from all solutions of the homogeneous equation. Thus our final assumption is

Y(t) = At2 sin t + Bt2 cos t.

Next, we differentiate Y( t) four times, substitute into the differential equation (4), and collect terms, obtaining finally

—8 A sin t — 8 B cos t = 3sin t — 5 cos t.

Thus A = — |, B = 5, and the particular solution of Eq. (4) is

Y(t) = —8t2sin t + 5t2cos t. (7)

If g(t) is a sum of several terms, it is often easier in practice to compute separately the particular solution corresponding to each term in g(t). As for the second order equation, 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.

Find a particular solution of

/" — 4y = t + 3 cos t + e—2t. (8)

First we solve the homogeneous equation. The characteristic equation is r3 — 4r = 0, and the roots are 0, ±2; hence

yc (t) = Cj + c2 e2t + c3e—2t.

Chapter 4. Higher Order Linear Equations

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

Ó" — 4Ó = t, Ó" — 4/ = 3 cos t, Ó" — 4/ = e~2t.

Our initial choice for a particular solution Yl(t) of the first equation is A0t + Al, but since a constant is a solution of the homogeneous equation, we multiply by t. Thus

Y() = t (V + A1).

For the second equation we choose

Y2(t) = Bcos t + C sin t,

and there is no need to modify this initial choice since cos t and sin t are not solutions of the homogeneous equation. Finally, for the third equation, since e~2t is a solution of the homogeneous equation, we assume that

Y3(t) = Ete—2t.

The constants are determined by substituting into the individual differential equations; they are A0 = —8, A1 = 0, B = 0, C =— |, and E = 1. Hence a particular solution of Eq. (8) is

Y(t) = — t2 — 3 sin t + 8 te~2t. (9)

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 for Y(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 Problems 1 through 8 determine the general solution of the given differential equation.

1. Ó" — ó" — Ó + Ó = 2e—t + 3 2. /v — Ó = 3t + cos t

3. /" + Ó' + Ó + y = e—t + 4t 4. Ó'' — Ó = 2sint

5. yv — 4 Ó = t2 + e 6. yv + 2 y"+ y = 3 + cos2t

7. yvi + Ó" = t 8. /v + Ó" = sin2t

In each of Problems 9 through 12 find the solution of the given initial value problem. Then plot a graph of the solution.

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