# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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EXAMPLE

1

Find a particular solution of

/- 3?- 4y = 3e2t. (9)

We seek a function Y such that the combination Y"(t) - 3Y'(t) - 4 Y(t) is equal to 3e2t. Since the exponential function reproduces itself through differentiation, the most plausible way to achieve the desired result is to assume that Y(t) is some multiple of e2t, that is,

Y(t) = Ae2t,

where the coefficient A is yet to be determined. To find A we calculate Y'(t) = 2 Ae2t, Y"(t) = 4 Ae2t, and substitute for y, y, and y" in Eq. (9). We obtain

(4 A - 6 A - 4 A)e2t = 3e2t.

Hence -6 Ae2t must equal 3e2t, so A = —1/2. Thus aparticular solution is

Y (t) = -1 e2t. (10)

172

Chapter 3. Second Order Linear Equations

EXAMPLE

2

Find a particular solution of

/- 3/- 4y = 2 sin t. (11)

By analogy with Example 1, let us first assume that Y(t) = A sin t, where A is a constant to be determined. On substituting in Eq. (11) and rearranging the terms, we obtain

—5 A sin t — 3 A cos t = 2sin t,

or

(2 + 5 A) sin t + 3 A cos t = 0. (12)

The functions sin t and cos t are linearly independent, so Eq. (12) can hold on an interval only if the coefficients 2 + 5 A and 3 A are both zero. These contradictory requirements mean that there is no choice of the constant A that makes Eq. (12) true for all t. Thus we conclude that our assumption concerning Y (t) is inadequate. The appearance of the cosine term in Eq. (12) suggests that we modify our original assumption to include a cosine term in Y(t), that is,

Y( t) = A sin t + B cos t, where A and B are to be determined. Then

Y'(t) = A cos t — B sin t, Y"(t) = — A sin t — B cos t.

By substituting these expressions for y, y, and y" in Eq. (11) and collecting terms, we obtain

(—A + 3 B — 4 A) sin t + (— B — 3 A — 4B) cos t = 2sin t. (13)

To satisfy Eq. (13) we must match the coefficients of sin t and cos t on each side of the equation; thus A and B must satisfy the equations

—5 A + 3 B = 2, —3A — 5 B = 0.

Hence A = —5/17 and B = 3/17, so a particular solution of Eq. (11) is

Y(t) = —17 sin t + 17 cos t.

The method illustrated in the preceding examples can also be used when the right side of the equation is a polynomial. Thus, to find a particular solution of

/ — 3 / — 4y = 4t2 — 1, (14)

we initially assume that Y (t) is a polynomial of the same degree as the nonhomogeneous term, that is, Y(t) = At2 + Bt + C.

To summarize our conclusions up to this point: if the nonhomogeneous term g(t) in Eq. (1) is an exponential function eat, then assume that Y(t) is proportional to the same exponential function; if g(t) is sin /31 or cos j31, then assume that Y(t) is a linear combination of sin 31 and cos 31; if g(t) is a polynomial, then assume that Y(t) is a polynomial of like degree. The same principle extends to the case where g(t) is a product of any two, or all three, of these types of functions, as the next example illustrates.

3.6 Nonhomogeneous Equations; Method of Undetermined Coefficients

173

EXAMPLE

3

EXAMPLE

4

Find a particular solution of

y" — 3/ — 4y =—8^ cos2t. (15)

In this case we assume that Y(t) is the product of el and a linear combination of cos 2t and sin 2t, that is,

Y (t) = Ae' cos2t + Be' sin2t.

The algebra is more tedious in this example, but it follows that

Y'(t) = (A + 2B)el cos2t + (—2A + B)el sin2t

and

Y"(t) = (—3 A + 4B)el cos 2t + (—4 A — 3B)el sin2t.

By substituting these expressions in Eq. (15), we find that A and B must satisfy 10 A + 2B = 8, 2 A — 10B = 0.

Hence A = 10/13 and B = 2/13; therefore a particular solution of Eq. (15) is

Y(t) = 10el cos2t + 13el sin2t.

Now suppose that g( t) is the sum of two terms, g(t) = g1(t) + g2(t), and suppose that Y1 and Y2 are solutions of the equations

ay" + by + cy = g1(t) (16)

and

a/'+ by + cy = g2(t), (17)

respectively. Then Y1 + Y2 is a solution of the equation

ay + by + cy = g(t). (18)

To prove this statement, substitute Y1(t) + Y2(t) for y in Eq. (18) and make use

of Eqs. (16) and (17). A similar conclusion holds if g( t) is the sum of any finite

number of terms. The practical significance of this result is that for an equation whose nonhomogeneous function g( t) can be expressed as a sum, one can consider instead several simpler equations and then add the results together. The following example is an illustration of this procedure.

Find a particular solution of

y — 3y — 4y = 3e2t + 2sin t — 8e( cos 2t. (19)

By splitting up the right side of Eq. (19), we obtain the three equations

/ — 3/ — 4 y = 3e2t, y — 3y — 4y = 2 sin t,

and

y — 3/ — 4y = —8e^ cos2t.

174

Chapter 3. Second Order Linear Equations

I

EXAMPLE

5

Solutions of these three equations have been found in Examples 1,2, and 3, respectively. Therefore a particular solution of Eq. (19) is their sum, namely,

Y(t) = — 2e2t + 17 cos t — 17 sin t + 10 el cos 2t + Hel sin2t.

The procedure illustrated in these examples enables us to solve a fairly large class of problems in a reasonably efficient manner. However, there is one difficulty that sometimes occurs. The next example illustrates how it arises.

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