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

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EXAMPLE

1

Find a particular solution of

/- 3/- 4ó = 3e2t. (9)

We seek a function Y such that the combination Y" (t) - 3 Y'( t) - 4Y(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, /, 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) = - e2t. (10)

172

Chapter 3. Second Order Linear Equations

EXAMPLE

2

Find a particular solution of

ó"- 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 ofthe 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 ó, Ó, and ó" 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, -3 A - 5 B = 0.

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

Y(t) = —(7 sin t + (7 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 â t or cos â t, then assume that Y(t) is a linear combination of sin â t and cos â t; 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 =— 8º* cos2t. (15)

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

Y (t) = Ael cos2t + Be1 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 ofEq. (15) is

Y(t) = (0el cos2t + (3el sin2t.

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

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

and

àÓ' + by + cy = g2(t), (17)

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

àÓ + by + cy = g(t). (18)

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