Question

(1 point) Use the Laplace transform to solve the following initial value problem: x' = 5x + 3y, y = -2x +36 x(0) = 0, y0) = 0 Let X(s) = L{x(t)}, and Ys) = L{y(t)}. Find the expressions you obtain by taking the Laplace transform of both differential equations and solving for YS) and X (s): X(S) = Y(s) = Find the partial fraction decomposition of X(s) and Y(s) and their inverse Laplace transforms to find the solution of the system of Des: 355 x(t) = yt) =

Please help solving all parts to this problem and show steps.

Answer 1

given system of differential eqn 16 se' els = 58 PBy ? y' = -28+ e3t with 260)= 0 Y(0) = 0 Jer XCS)= Lfae Cty and YCB)2 Lf YCH} taking laplace fernsform of all we get cfae'cti}= 1{58434} - LP Y CH} = {-2x+8t} we know if bffCh} = FOS) then {{f\CH2=529fC+}-f L{XCH} = 95 X(s) - 2e (0) Elf464)}= 7C55 - 960) SXCS) - 200) = 28583 +28 3y} SY(5) - yco) = 1 {-283+ 31} SXCS) – 0 = 5482Ct)} + 3284€+)} SY18 -0 = -2488434ed (al feat}ista 5xc5) = 5x0) +3705) SYC) -2xes + $-3

XC5) (5-5) – 37C5) = 0 —ci) 2x05) + SYC5) = Šog C which is system of linear equation in (s) and Y(S) :: We solve this system of lineoro equation using coamers sale 3-5 На“) е Here Dx = 180 m3 / Jess) of s 12 S-3 = ba (55)5 +6 Dx= 0(s) - 3 - 2 aby Songs

Xos) CS-3) SCS-5 + = (-3){$cs-5300) (8-55+6) (59) T Xcs) - 1933.6+(5-1)(82) (8-3) ] Rope 3 YOS) = (": 3-5546=(5-2)(5-)) pels (5-5) CS-3) (3-5)5+6 (5-5) CS-3)(3-5)5+6) (9-5) (5-3)(3-5516) (5-5) CS-3) (5-3)(3-2) T YCS) s-5 · Here XC5) = 75-3757 CS-2and yes) a 5-3) 65-3) S_74 we have to find laplace in verse of XCs) and (8) for this we use postical fraction method . We all find posticul fruction decomposition of xcs) and YCS) considers xes) = 632 (5-1)(62) = conoscesas, consideren e he sy . consider la (5-3)(3-3) (5-2) A (5-3) (3-2)

iz ŠACS-3) (5-23 F B CS-3; + CCS-332- a put 822 woog in @ we get 1o CCD? = TC= put sa 3 in ③ we get TEB01) 'Barl put 320 in we get 1: AC-3687 B (-34 + (3)? la 6A * -2B +90 - 6A - 20D + g(1) 6A = 1+2-9 = = 6A = -6 IAI AS-I) BEL ecst ***]0-968-25 = ishin te nga ta's af stoca} = tf * cipela o 1. We know He } cat silte stata est et Also we know frost shifting en bos RCASESA-tetren : bele ." IF LS FCs)} = fct) then El{ FCS-4)= et fch} ... Here Hf} = t ::41 6 bajr} = catt Les segal= 2tz for a=3

2. L'{ colocacoy }= 237 pettet = pat(tt) + 2 :, L48 Xcs)} = H 3693054 %= 3 (2+C144) + 3.44 E 2CH) = 33+(147) + 3 + 1 Now fer rss S-5 (5-5). (3-3)(3-4 (5-3) (5-3)} (S-2 We Hindi pestial decom position of ros) set ou t to there - 5-15 = A(5-3)(3-2) + B(5-2) + CCS-332 Put 322 in weget 3= M CC-² 1ca3] put 8=3 in 6 we get 2= BCID = B=2 put 320 in C Đ we get -5= AC-3) (-2) + B (-2) + c(-37² -5 = 6A P 26-31 + (3)C1) 53 GA - 4 p 24 6A = 64-275 SA = -28 A = -14

CS-2) CS-3) + Y00) = 6 *) to get .: 1fycosya allar } + 2 be} 43218f} Hif y(s)} = th 3 e 2test + Belt if yet = -14 3+ + 2te3te Bed sect) = Betclits toet yet = -14 34 24 37 4384 . are solutions of given systems of ten- diffean