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is a solution of the nonhomogeneous equation (1).
To determine u1 and u2 we need to substitute for yfrom 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 up 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 u1 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
/ = —2u1(t) sin2t + 2u2(t) cos2t + u'l(t) cos2t + u'2(t) sin2t. (5)
Keeping in mind the possibility of choosing a second condition on and u2, let us require the last two terms on the right side ofEq. (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.
3.7 Variation ofParameters
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" = -4uj(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[(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) = -u^t). (10)
Then, substituting for u'2(t) in Eq. (9) and simplifying, we obtain
,, ÷ 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 u1(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 j(t) = — 3sint + cl (13)
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 + cx cos 2t + c2 sin 2t,
y = 3sin t + 2 ln | csc t — cot 11 sin2t + cr cos2t + c2 sin2t. (15)
The terms in Eq. (15) involving the arbitrary constants cx and c2 are the general solution of the corresponding homogeneous equation, while the remaining terms are a particular solution ofthe nonhomogeneous equation (1). Therefore Eq. (15) is the general solution ofEq. (1).
In the preceding example the method of variation of parameters worked well in determining a particular solution, and hence the general solution, ofEq. (1). The next
Chapter 3. Second Order Linear Equations
question is whether this method can be applied effectively to an arbitrary equation. Therefore we consider
/ + p(t) Ó + 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
Óñ(t) = Ñ1Ó1(t) + c2y2(t) (17)
of the corresponding homogeneous equation
/ + p(t)y + 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 y().
The crucial idea, as illustrated in Example 1, is to replace the constants ñ1 and ñ2 in Eq. (17) by functions u 1(t) and u2(t), respectively; this gives
Ó = u() y() + U2(t) y2(t). (19)