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

ISBN 0-471-31999-6

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approach. Observe also that the homogeneous equation corresponding to Eq. (1) is

/' + 4 y = 0, (2)

and that the general solution of Eq. (2) is

yc(t) = c1 cos 2t + c2 sin 2t. (3)

The basic idea in the method of variation of parameters is to replace the constants c1 and c2 in Eq. (3) by functions u1(t) and u2(t), respectively, and then to determine these functions so that the resulting expression

y = u1(t) cos2t + u2(t) sin2t (4)

is a solution of the nonhomogeneous equation (1).

To determine u1 and u2 we need to substitute for y from Eq. (4) in Eq. (1). However,

even without carrying out this substitution, we can anticipate that the result will be a

single equation involving some combination of u 1, u2, and their first two derivatives. Since there is only one equation and two unknown functions, we can expect that there are many possible choices of u 1 and u2 that will meet our needs. Alternatively, we may be able to impose a second condition of our own choosing, thereby obtaining two equations for the two unknown functions u 1 and u2. We will soon show (following Lagrange) that it is possible to choose this second condition in a way that makes the computation markedly more efficient.

Returning now to Eq. (4), we differentiate it and rearrange the terms, thereby obtaining

y = —2u1(t) sin2t + 2u2(t) cos2t + u'1(t) cos2t + u'2(t) sin2t. (5)

Keeping in mind the possibility of choosing a second condition on u1 and u2, let us require the last two terms on the right side of Eq. (5) to be zero; that is, we require that

u[(t) cos2t + u'2(t) sin2t = 0. (6)

It then follows from Eq. (5) that

y = —2ul(t) sin2t + 2u2(t) cos2t.

(7)

3.7 Variation ofParameters

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Although the ultimate effect of the condition (6) is not yet clear, at the very least it has simplified the expression for /. Further, by differentiating Eq. (7), we obtain

y" = —4u 1(t) cos2t — 4u2(t) sin2t — 2u[(t) sin2t + 2u'2(t) cos2t. (8)

Then, substituting for y and y" in Eq. (1) from Eqs. (4) and (8), respectively, we find that u1 and u2 must satisfy

—2u'1(t) sin2t + 2u'2(t) cos2t = 3 csc t. (9)

Summarizing our results to this point, we want to choose u1 and u2 so as to satisfy Eqs. (6) and (9). These equations can be viewed as a pair of linear algebraic equations for the two unknown quantities u[(t) and u'2(t). Equations (6) and (9) can be solved in various ways. For example, solving Eq. (6) for u'2(t), we have

u'2(t) = -u1(t). (10)

sin2t

Then, substituting for u'2(t) in Eq. (9) and simplifying, we obtain

,, s 3csc t sin2t „ u 1(t) =-----------2--------= —3 cos t. (11)

Further, putting this expression for u[(t) back in Eq. (10) and using the double angle formulas, we find that

3cost cos2t 3(1 — 2sin21) 3 .

u2(t) =---------^T.-------=----------------= - csc t — 3 sin t. (12)

sin2t 2sint 2

Having obtained u[(t) and u'2(t), the next step is to integrate so as to obtain u 1(t) and u2 (t). The result is

u 1(t) = — 3sint + c1 (13)

and

u2(t) = 3 ln | csc t — cot 11+ 3 cos t + c2. (14)

Finally, on substituting these expressions in Eq. (4), we have

y =— 3 sin t cos2t + 3 ln | csc t — cot 11 sin2t + 3 cos t sin2t + c1 cos 2t + c2 sin 2t,

or

y = 3sin t + 2 ln | csc t — cot 11 sin2t + c1 cos2t + c2 sin2t. (15)

The terms in Eq. (15) involving the arbitrary constants c1 and c2 are the general solution of the corresponding homogeneous equation, while the remaining terms are a particular solution of the nonhomogeneous equation (1). Therefore Eq. (15) is the general solution of Eq. (1).

In the preceding example the method of variation of parameters worked well in determining a particular solution, and hence the general solution, of Eq. (1). The next

182

Chapter 3. Second Order Linear Equations

question is whether this method can be applied effectively to an arbitrary equation. Therefore we consider

/' + p(t) / + q (t) y = g(t), (16)

where p, q, and g are given continuous functions. As a starting point, we assume that we know the general solution

yc(t) = c1 y1(t) + c2y2(t) (17)

of the corresponding homogeneous equation

f + P(t)/ + q (t) y = 0. (18)

This is a major assumption because so far we have shown how to solve Eq. (18) only if it has constant coefficients. If Eq. (18) has coefficients that depend on t, then usually the methods described in Chapter 5 must be used to obtain yc(t).

The crucial idea, as illustrated in Example 1, is to replace the constants c1 and c2 in

Eq. (17) by functions u 1(t) and u2(t), respectively; this gives

y = u() y() + u2(t) y2(t). (19)

Then we try to determine u1 (t) and u2(t) so that the expression in Eq. (19) is a solution of the nonhomogeneous equation (16) rather than the homogeneous equation (18). Thus we differentiate Eq. (19), obtaining

y = u\(t) y1 (t) + u^t) y1 (t) + u2(t) y2 (t) + u 2(t) y2 (t). (20)

As in Example 1, we now set the terms involving u[(t) and u'2(t) in Eq. (20) equal to zero; that is, we require that

u[(t) y() + u2(t) y2(t) = 0. (21)

Then, from Eq. (20), we have

y = Ul(^) y1 (t) + u2(t) /2 (t). (22)

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