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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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4 2!
2. We have --------- = 2---------- and thus the inverse Laplace
/ 3 / - \ 2 + 1
(s-1) (s-1)
2t
transform is 2t e , using line 11.
3s 3s 9/5 6/5
4. We have --------- = ------------- = ------ + ---- using partial
s2-s-6 (s-3)(s+2) s-3 s+2
3t -2t
fractions. Thus (9/5)e + (6/5)e is the inverse
transform, from line 2.
2s+1 2s+1 2(s-1) 3
7. We have --------- = ----------- = ----------- + -----------, where
s2-2s+2 (s-1) 2+1 (s-1) 2+1 (s-1) 2+1
106
Section 6.2
we first used the concept of completing the square (in the denominator) and then added and subtracted appropriately to put the numerator in the desired form. Lines 9 and 10 may now be used to find the desired result.
In each of the Problems 11 through 23 it is assumed that the
I.V.P. has a solution y = f(t) which, with its first two derivatives, satisfies the conditions of the Corollary to Theorem 6.2.1.
II. Take the Laplace transform of the D.E., using Eq.(1) and Eq.(2), to get
s2Y(s) - sy(0) - y'(0) - [sY(s) - y(0)] - 6Y(s) = 0.
Using the I.C. and solving for Y(s) we obtain s-2
Y(s) = -------. Following the pattern of Eq.(12) we have
s2-s-6
s-2 a b a(s-3)+b(s+2)
------ = ----- + ----- = ---------------. Equating like
s2-s-6 s+2 s-3 (s+2)(s-3)
powers in the numerators we find a+b = 1 and -3a + 2b = -2. Thus a = 4/5 and b = 1/5 and 4+5 1/5
Y(s) = ---- + ----, which yields the desired solution
s+2 s-3
using Table 6.2.1.
14. Taking the Laplace transform we have s2Y(s) - sy(0) -y'(0) -
4[sY(s)-y(0)] + 4Y(s) = 0. Using the I.C. and solving for
s-3
Y(s) we find Y(s) = -----------. Since the denominator is a
2
s -4s+4
s-3
perfect square, the partial fraction form is
2
s -4s+4
ab
------ + ----. Solving for a and b, as shown in examples of
(s-2)2 s-2
this section or in Problem 11, we find a = -1 and b = 1.
11
Thus Y(s) = ----- - ---------, from which we find
s-2 (s-2)2
y(t) = e2t - te2t (lines 2 and 11 in Table 6.2.1).
2s-4 2s-4 2(s-1) 2
15. Note that Y(s) = ------------=----=------------------.
s2-2s-2 (s-1) 2-3 (s-1) 2-3 (s-1) 2-3
Three formulas in Table 6.2.1 are now needed: F(s-c) in
line 14 in conjunction with the ones for coshat and
sinhat, lines 7 and 8.
Section 6.2
107
17. The Laplace transform of the D.E. is
s4Y(s) - s3y(0) - s2y'(0) - sy"(0) - y"'(0) - 4[s3Y(s)-s2y(0)
-sy'(0) - y"(0)] + 6[s2Y(s) - sy(0) - y'(0)] - 4[sY(s) - y(0)] +Y(s) = 0. Using the I.C. and solving for Y(s) we find
2
s2 - 4s + 7
Y(s) = ----------------. The correct partial fraction
432
s -4s +6s -4s+1
a b c d
form for this is
(s-1) 4 (s-1) 3 (s-1) 2 s 1
Setting this equal to Y(s) above and equating the
22 numerators we have s -4s+7 = a + b(s-1) + c(s-1) +
d(s-1)3. Solving for a,b,c, and d and use of Table 6.2.1
yields the desired solution.
20. The Laplace transform of the D.E. is
s2Y(s) - sy(0) - y'(0) + w 2Y(s) = s/(s2+4). Applying the
2 2 2
I.C. and solving for Y(s) we get Y(s) = s/[(s +4)(s +w )] 22
+ s/(s +w ). Decomposing the first term by partial
fractions we have
s s s
Y(s) =------------------------------------+ --------
(w2-4)(s2+4) (w2-4)(s2+w2) s2+w2
2
2 -1 (w -5)s s
= (w2-4) 1[---------- + ------].
2 2 2
s2 +w 2 s2 +4
Then, using Table 6.1.2, we have
2 -1 2 y = (w -4) [(w -5)coswt + cos2t].
2
22. Solving for Y(s) we find Y(s) = 1/[(s-1) + 1] +
2
1/(s+1)[(s-1) + 1]. Using partial fractions on the
second term we obtain Y(s) = 1/[(s-1) 2 + 1] + {1/(s + 1) - (s-3)/[(s-1) 2 + 1]}/5 = (1/5){(s + 1) -1-(s-1)[(s-1) 2 + 1] -1 + 7[(s-1) 2 + 1] -1}. Hence, y = (1/5)(e-t - etcost + 7etsint).
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