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

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4 2!

2. We have --------- = 2---------- and thus the inverse Laplace

/ ³ ÷ 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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