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

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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5. Find a particular solution Y (t) for each of the subproblems. Then the sum Y1(t) +
+ Yn (t) is a particular solution of the full nonhomogeneous equation (23).
6. Form the sum of the general solution of the homogeneous equation (step 1) and the particular solution of the nonhomogeneous equation (step 5). This is the general solution of the nonhomogeneous equation.
7. Use the initial conditions to determine the values of the arbitrary constants remaining in the general solution.
TABLE 3.6.1 The Particular Solution of ay/' + b/ + cy = gt ( t)
g, (t) Yt (t)
Pn (t) = aotn + a^-1 +.. + an ts(A0tn + Aj tn-1 + ... + An)
ts(A0tn + Aj tn-1 + ... + An)eat
) ts[(A0^ + Ajtn-1 +-----+ An)eat cos t
e + (B0tn + + ... + Bn )eat sin t]
t
cs
n
s
^
t
Notes. Here s is the smallest nonnegative integer (s = 0, 1, or 2) that will ensure that no term in Yj (t) is a solution of the corresponding homogeneous equation. Equivalently, for the three cases, s is the number of times 0 is a root of the characteristic equation, a is a root of the characteristic equation, and a + ip is a root of the characteristic equation, respectively.
176
Chapter 3. Second Order Linear Equations
For some problems this entire procedure is easy to carry out by hand, but in many cases it requires considerable algebra. Once you understand clearly how the method works, a computer algebra system can be of great assistance in executing the details.
The method of undetermined coefficients is self-correcting in the sense that if one assumes too little for Y(t), then a contradiction is soon reached that usually points the way to the modification that is needed in the assumed form. On the other hand, if one assumes too many terms, then some unnecessary work is done and some coefficients turn out to be zero, but at least the correct answer is obtained.
Proof of the Method of Undetermined Coefficients. In the preceding discussion we have described the method of undetermined coefficients on the basis of several examples. To prove that the procedure always works as stated, we now give a general argument, in which we consider several cases corresponding to different forms for the nonhomogeneous term g( t).
g(t) = Pn (t) = a0 tn + a1 tn-1 + + an. In this case Eq. (23) becomes
ay + by + cy = a0tn + a^-1 + ... + an. (24)
To obtain a particular solution we assume that
Y(t) = A0tn + A^-1 + . . . + An-/ + An_!f + An. (25)
Substituting in Eq. (24), we obtain
a[n(n - 1) A0tn 2 + ... + 2An-2] + b(nA0tn 1 + ... + An-i)
+ c( A0tn + A1tn 1 + ... + An) = a0tn + ... + an (26) Equating the coefficients of like powers of t gives
cA0 = a0, cA1 + nbA0 = ap
cAn + bAn-1 + 2aAn-2 = an
Provided that c = 0, the solution of the first equation is A0 = a0/ c, and the remaining equations determine A1t, An successively. If c = 0, but b = 0, then the polynomial on the left side of Eq. (26) is of degree n - 1, and we cannot satisfy Eq. (26). To be sure that aY"(t) + bY'(t) is a polynomial of degree n, we must choose Y(t) to be a polynomial of degree n + 1. Hence we assume that
Y (t) = t (A0tn + ... + An).
There is no constant term in this expression for Y(t), but there is no need to include such a term since a constant is a solution of the homogeneous equation when c = 0. Since b = 0, we have A0 = a0/b(n + 1), and the other coefficients A1t..., An can be determined similarly. If both c and b are zero, we assume that
Y (t) = 12( A0tn + ... + An).
The term aY''(t) gives rise to a term of degree n, and we can proceed as before. Again the constant and linear terms in Y(t) are omitted, since in this case they are both solutions of the homogeneous equation.
3.6 Nonhomogeneous Equations; Method of Undetermined Coefficients
177
g(t) = e?t Pn (t). The problem of determining a particular solution of
a/ + by + cy = eatPn (t) (27)
can be reduced to the preceding case by a substitution. Let
Y (t) = eatu(t);
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