Question

Consider the following initial value problem, representing the response of an undamped oscillator subject to the ramp loading applied force g(t): 0<t<2, y' + 25y = g(t), y(0) = 0, y(0) = -2, g(t) = 2<t<6, otherwise. 0 t - 2 2 2 In the following parts, use h(t – c) for the Heaviside function he(t) when necessary. a. First, compute the Laplace transform of g(t). L{f(t)}(s) b. Next, take the Laplace transform of the left-hand-side of the differential equation, set it equal to your answer from Part a. and solve for L{y(t)} L{y(t)}(s) = c. Lastly, take the inverse Laplace transform to compute y(t). g(t) = Instructions

Answer 1

solution :

Given g(t) = ft-2 ostc2, 2 <t<6 otherwise a); L { 9(e)} (S) = sest scoot oest dt + s (m2). ES stedt + jest Ej2 slidt 2 6 SE -1) SE dt+ Jé, dt 2 6 L {9CE)y cst estatal st 1.e att st e dt 2 2 0 fateve As we knoco that filmetdie = d.cre-Sockelta del take got est est de So. az el dt - -st 2 -S Il -6 -65 -65 - 23 -25 sers 름이 + - Le =é 28 ( n -)-( tandems Now take it - 2 Stést at .1 ģē --25 + 3 2 825 2 -65 252 ższērs How take -25 65 -GS е -S te -25 е s 2 S S Sestat: The De Now take sest, ata lease Therefore from eano +{900)} (5) = s -6S e -65 6 + 1/3 6 Š 25 25 +1 2Sم . 3 e 6s , -65 e 25² # S 25² eln f 760 S . :-/L (364)](5) ē 25 -65 e S 282 ës 252

Given y"+25y = 9 (t) 2 40020; 4(0)=-4 (b) Taking Laplace on both sides of the initial value Problem L -25 e -65 (4) + 25 (4) = L(Ct) :) SL (4(E)) -54603-4'60) +25L (YC#))= źS2 =>15 5+25) 44C+J) –0-6-2) = 725 (5'425) L L (4(t)) = e 6 كاتم ل 28² -65 -e ~65 25² S 282 =-2+ e as -65 282 => L [4CŁY(s) = -G'S e 2 2S S -2 (5+25) të 25 abs - es 256425) 5(5425) 2548425) L {4C+3}(3) = 1 -a + e25 -65 e -65 e - (5725 282 S 282 (c) Nocu EL 4 (t) = {'{-(46+)) (s)} {"Cs) +"(686-- Ver pintar (3313) = - the take Noco 4 :) How take 5 S452 - (5425)-5² $2(5+25) 255(5425) sinst 5 1 ( 25 2 3425 fontos ( 64 Pers)= $ [1" (+) -1"( .#3 (8-sist) ولی --) L (as? (425) re?s 50 1-2)-sin5(7-2)) h(0-2) so (4t-2)=sin(57-10) h(+-2) S Now take 1 +29)-2 S (5425) 255 (s +25) 51S S (5425) So, now F'( Sistas) = 35 (296)-1(10) 75503

=) Ľ ( S(5725) Friss is a I (W=1 El (S211%) = Cosbe Cosslagher NOW. 2-1 e EC) 25 (125) (4-0)-simft. ) h(+6) 50 i'ceas F(s)) = f(t-a) b(ta) where F(s)L(P(+)) 2.4Ct) 50 betes 50 sin5t + —2) -- 0 (54-10)) h(+-2) (1.- C056-30)) - [ct-6) - Sinßt=30 -Sint-30]ht- :: 466) czasinst to y Ct -->-s'obt-1) nct-2)] of4-6) - Sinist_29) het hy Eco-cosasto 50 h(t-6)