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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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Substituting for d (t - t0) from Eq. (4), we obtain
tn+t
1
C{dr(t - t0)} = I e~st dt = - e-st
t0
2 t 2s
t-t0+r t=0-
= e-st0 (esx - e-sx)
2s
or
sinh s
C{dx(t - t0)}=----------------e~st0. (12)
^ s
The quotient (sinh s^/s is indeterminate as 0, but its limit can be evaluated by LHospitals rule. We obtain
sinh s s cosh s
lim-----------= lim---------------= 1.
0 s 0 s
Then from Eq. (11) it follows that
L{S(t - 0)} = e-st0. (13)
2Paul A. M. Dirac (1902-1984), English mathematical physicist, received his Ph.D. from Cambridge in 1926 and was professor of mathematics there until 1969. He was awarded the Nobel prize in 1933 (with Erwin Schrodinger) for fundamental work in quantum mechanics. His most celebrated result was the relativistic equation for the electron, published in 1928. From this equation he predicted the existence of an anti-electron, or positron, which was first observed in 1932. Following his retirement from Cambridge, Dirac moved to the United States and held a research professorship at Florida State University.
6.5 Impulse Functions
Equation (13) defines C{8(t t0)} for any t0 > 0. We extend this result, to allow t0 to
be zero, by letting t0 ^ 0 on the right side of Eq. (13); thus
?{5(t)} = lim est0 = 1. (14)
t0 ^0
In a similar way it is possible to define the integral of the product of the delta function and any continuous function f. We have
/00
S(t 0 f(t) dt = lim / dx(t 0 f (t) dt. (15)
r^0 J o
Using the definition (4) of dT (t) and the mean value theorem for integrals, we find that
f0 1 f (0+t dr (t t0) f (t) dt =\ f (t) dt
J-oo 2 t Jl-t
= 2 t f(t*) = f(t*),
2
where t0 t < t* < t0 + t. Hence, t* ^ t0 as t ^ 0, and it follows from Eq. (15) that 0 0 0
*
S(t 0 f(t) dt = f(t0). (16)
f
J
It is often convenient to introduce the delta function when working with impulse problems, and to operate formally on it as though it were a function of the ordinary kind. This is illustrated in the example below. It is important to realize, however, that the ultimate justification of such procedures must rest on a careful analysis of the limiting operations involved. Such a rigorous mathematical theory has been developed, but we do not discuss it here.
Find the solution of the initial value problem
2/ + / + 2 y = S(t 5), (17)
y(0) = 0, (0) = 0. (18)
This initial value problem arises from the study of the same electrical circuit or me-
chanical oscillator as in Example 1 of Section 6.4. The only difference is in the forcing term.
To solve the given problem we take the Laplace transform of the differential equation and use the initial conditions, obtaining
(2s2 + s + 2) Y (s) = e~5s.
Thus
e5s e5s
Y(s) = 2------------------= ^----------------------------. (19)
2s + s + 2 2 (s + 4) +
By Theorem 6.3.2 or from line 9 of Table 6.2.1
H (T+FTT* e-"4sm# ' (20)
328
Chapter 6. The Laplace Transform
Hence, by Theorem 6.3.1, we have
y = ?~1{Y(s)} = -L u5(t)e-(ti)/4sm>f (t 5), (21)
which is the formal solution of the given problem. It is also possible to write y in the form
f 0, t < 5,
y =j -= e(t=)/4 sin -= (t 5), t > 5. (22)
The graph ofEq. (22) is shown in Figure 6.5.3. Since the initial conditions at t = 0 are homogeneous, and there is no external excitation until t = 5, there is no response
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