Question

Homework Set 5 f(t) F(S) Section 4.1: Apply the definition to directly find the Laplace transforms of the given functions. (s > 0) 1 (s > 0) S- 1. Kt) = 12 2. f = 23t+1 Use transforms from the Table (op right) to find the Laplace transforms of the given functions. t" ( n20) (s > 0) r(a + 1) 1a (a > -1) (s > 0) 5+1 3. f(t) = VE +8t 4. f(t) = sin(2tcos(2t) Use the transforms from the Table (on right) to find the inverse Laplace transforms of the given functions. (s > 0) S-a cos kt ali -13 -13 -E -3 -1 (s > 0) 5. F(s) = 6. F(s) 1-2 sin kt (s > 0) S2 Section 4.24 cosh kt (s > Ik) Use Laplace transforms to solve the following initial value problems. k sinh kt (s > |k 7. x" - x' - 2x = 0; x(0) = 0,x'(0) = 2 8. X" + x = cos(3t); x(0) = 1, x'(0) = 0 ut -a (s > 0) S Section 4.3: Apply the translation theorem to find the Laplace transforms or inverse Laplace transform of the given functions. Note that I (a + 1) = a! r-vo 9. f(t) = tent 10. F(s) = S-1 (5+1) (You do not need partial fractions here.) Use partial fractions to find the inverse Laplace transforms of the given functions. 5-25 11. F(S) = $2+75+10 12. F(S) = (52 +5-6) Use Laplace transforms to solve the initial value problems. 13. X 6x' + 25x = 0; x(0) = 2 U = 3 14. x" +4x13x = te:O) = 0,X'O) = 2 15. x(4) + 2x" + x e2%, x(0) = x'O)=x"(0) = x(3)(0) = 0 Solve the problem: 16. Consider a mass-and-spring system with m = 1, k = 9, c = 0, and F(t) = 6 cos(3t). Use the given Laplace transforms to solve the initial value problem with x(0) = x'(0) = 0. 1 (s2 + k232) t sin(kt) --+62 +222}=2*** +7(6+237}= 2## (sin(kt) – kt cos(kt)

Answer 1

g f(t)= the 4 tt o 4 4. 2 Taking laplace shipting the 55 How, we know that them Fes-a) L} stjy = Fes) Lheat fet)} Lent.tt} t 4 24 is-15)5 (5-0) 5 FIS) 10 3-) (5+172 FES) = sti-2 cst12 Take Invesse Laplace 2 S 2 52 11 et [' 3) [('} *}-1973) et (1 - 2+] .t e fet) =

use partial Fri) 5-25 s²+75+10 traction Fes) + B (5+5) (5+2) [(3²+75+10) (5+5) (3+2) A 5 + 4 A = 5-28 st5 J 3 15=-2 3 15 10 =-5 5-25 s +2 - 3 15=-5 3 Fesz Б sts st2 Take Inzesse lapla ale -2t fit) 3 è 5t 5 e Fes) = (5²75-6) 2 [(5+3)(3-2)]2 1 (5 +372 (5-2)2 By partial fraction, + D 4 f B (1-2) (5-2R (8 +872 (5+3)

I= A (5+3)(5-2)} + B (5-2) + ((S-2) (s+372 +D (5+3)2 Put s = -3 1/25 B (+25) - D - Puts=2 D(25) /25 put s=o 1=12A + 4 B - 18 ctg D = 124 + 4 - 180 +q 25 1 - P2A-180 - 1-13 25 12A-180 12 2A-3 c = 2 2 0 put S=1, 1 = 4 A + B - 16C +16D + 16 = 4 A 166 + 25 4 A-) 6 C = -17 4 A-l6c = 8 2 / at A-4c=2 28 By O 2 A-30 z 2A - 80 z 2/5 5c = - 2 23 125

A = 2 25 + 4C 8 A 긁 25 125 A = 2 125 25 FIS 2 2/25 2/125 (5-2) + t 2. st3 (5+3)² (5-2)2 Take Hace Lapla Inverse .t-2 ezt + I ett 25 -3t + 1 -3t e fit) = 2 e 2 125 125 25 ten 24 - 3t e 25 = [t-}] با او [ 233 لا tt 2 t