Question

Part It (page 1 of 3) X + kalamump.edu.my/mod/quiz/attempt.php?attempt=1751228cmid=112740 Recent. My Courses All Courses Blogs Teachers Guide FAQ - VIDEO CONTEST ORDINARY DIFFERENTIAL EQUATIONS Dashboard My courses BUM2133Topic 14 Assignment Part 1 Question 1 Not yet answered Marked out of 6.00 Flag question Use Laplace transforms to solve the following simultaneous differential equation, dar = x(t)+3y(t) dt =-3y(t)+5 (1) =-5-() dt subject to the given initial conditions: x(0)=0, y(0)=0,-(0)= 70- Find z(t), y(t) and then x(t). Select one r(t) = -2(et +2e-*—3-34), y(t) = o[e-3 +e-St), z(t)=zpe=3 r(t)-olette-4-2-34), y(t) = o(ex-e-5) =(t)=zpe=100 *r(t) =-2-+ +e-U+e), y(t) = o[e-34-4-34), 3(t)=300-54 Pol-de-1+e3+3), y(t) = - -st-e-5t) =(t) = get r/t) (el +20+ 0 –34–3), y(t) ==o[e-#-e-5) =(t)=20e-6 of-4e-++e-8f +3#), y(t) =30(-e-14), 2(t)=zpe- Antigament Part

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Answer 1

AGF NO. - at 21t) + 3y1+) =) x'lt) = q (t) +3814) - 3) + /(+) :) *') --38 + ) - $217) dy dt dz dt 52ct) z'lt) = -5z1t) alth - 21+) +3810) Taking L. T. ou leoth side. #10) = 70) = 0 Z Tol=20 S Xis) - X10) X75) + 3 15 ) S x 15 ) - X15) 3415 ) X(S)( 5 -1) - 3815) А. ylt) -371t) + SZ) L.T an leath side syis) -710) = - 3 yis) +5 ZTS) Sy(s) + 3 Y15) 5215) - yis) 15+3] $+3] =5215) yrs) (5+3) = 5205) B

zlit) = -5 zit) SZIS) - 210-5215) S21S)+5215) Z10) Zis sts) = 20 215) = Zo 5 +5 Taking invouplaplace to eens fororen 2/2 (s) = % 20% 75+5] 71+ -5t ño e freem the B NO the een putting the valué of zis) into equat YIS) (5+3) = 5 Zo 545 yes) (573) (5+5) 520 1 X = yis) $+3 SE5 403) = 15-545

Nowo turtce imuere Laplace Ytt) = Zo lo at est 2 into Now putting the value of y(s) equation A) s (T-SJ(9X T 3 XIS) T sts { 4 2 2. T T T X (S) وی ابر 20 x() S-1SF3) (-1) sts) ( - ) - ( 12 | () - () - HG) I 5+5 I T 4 6 1st5 (SJX S 2 (sts) +( 2 ( Inverse Laplace.

+ -3t -st 15 act) + Zo ( 12 Am ce 2 T5 Zo t st 11 3 e 3d the 모 12 is 7 3t = Zo e 3e +2e 24 t -zt 5 Zo a(t) = -57 + 20 e 3 e *] 8 & we have =) ft XL) Zetet & Zita st Zo e