Question

7.10.4 Question Help Use the method of Laplace transforms to solve the given initial value problem. Here, x' and y' denote differentiation with respect to t x' - 6x + 2y = sint x(O) = 0 29 7*-y' – 4y = cost y(0) = 0 Click the icon to view information on Laplace transforms. x(t) = y(t) = (Type exact answers in terms of e.)

DIFFERENTIAL EQUATIONS / Linear Algebra

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Answer 1

- Sint X(0) = 0 x'-6x + + 2y lleing the method og Laplace Transform - 6 X(S) + 27 (5) 2 simt SXIS) x 10) - S X(S) -0-6928) +271s) 1 52+1 I XIS) 15-6] + 2y (5) st A = cost 21 /2 x - x - 4y yo = 0 Laplace Thanjoom xis) -S415) + y(0) – 4 y(s) Al cost ] 29 = 2 29 yrs) -sy(s) – 4 405) S 5 + 1 29 B . s 62+1 X(S) – 415) (5+4) 2 Mullip lying @ by (5+4) and ® by 2 and addling them - XIS) [5-6) (5+4) + 2 65+4)ýcs) (+4) 52+1 25 s+1 (5+4) 29 X75) 2(5+1) Y(S) = Xis) /5-6) 18+8) +29 X75) - 3+1 + 25 3+1

28 + 4 st1 X(5) 1 (n=6) (5+4) +29) (cs-cb04 X(S) - ( 3st 4 2+1 (5-6) (5+4) +29 = 3 ste 2 I sé+45-65-24 +29 2 S”+1 1 xls) * (82 - 25 +5) (38+4) 9/s2+1) S +112 3/2-s Xis) 12 t 2 2 S +1 S-2 S +5 S t 3 2 S+ -) - sat1 2 5+1 4 (S-1)+22 (S-1) +24 Jaking imveuse Laplase =) cost t 3 t sinat 1/ / sint the et cost o-let 2 Smat X (f) cust+of Sint + et sinet - Of couet 2 4 Am

I 1 1 + sti lt) - sint confect et simit llow betting the value of xos) in pg") from a ( 35+4) (3-6) + 2 y(5) i 52-25+5°) (5+1) (35+4) (5-6) 2y(s) = 5+1 (52-25+5)/5*+1) 352-145-24 3+1 (52-25+5)(s’+1) I 18/2 -4 LIS 12 - $2+1 $2-25 +5 S 4 Il S-!+) 4 et 3+1 st! 2 52-25+5 $3-25+5 11 (5-1) 11 2 2+ s² +! 2 (S-1) 722 4(S-2) +22 (S-1)2+2 t -Il et sinat sind tll cost - 4 sint -!! e co824 + 2 et sinat # 3 8 4 11 + 2 11 S 4 + 2x2 t 2 forvente doplace 2 274) - 3 2 Ang