Question

QUESTION 1 use to following initial value problem (write fraction as After Laplace Transform transform the x" + 3x' + 2x=2e-t, x(0) = x'(0)=0, you should get X(s)= S-2 (S-2)/(5-4)(s+6) for (s-4)(s+6) -). Then, find x(t)= L-(x(s))= 5 -3t (write 5/6 by 6' ; e^{-3t} by e and sin(2t) or cos(3t) by sin(2t) or cos(3t)).

Answer 1

Given that Intial value problem x"1+3x)+2x=261 X(0)=160=0 applying both sides Laplace toemformntum LEX"(t)] + L {3xlt)} +2L{x} = 22 hét} LE0)= pdw-pro- .: L£$'(0)} = P062-06) L{001)] = 0CP) => SXC)-S X(0)-x.) +3[$Xcs)–Xo)] = 241@t} s”x(s)-5.0 -0+35 XCS) -3.0 = &. 5, X00)=x'0)=0 => SX() +35X(5). = 2 t => XC3) [5'+35] SH St X(s) = 2 (5+1) S(ST3) Q SC+1)5+3) XCS)= 2 SCStU(5+3) Lil t. 35 S+) XS) test [use partial Fraction applying both sides investe Laplace L! {X6)} = 2 [1553 - L'E 4,3 + 1/2 [ { sty} XE) L'{}}=1 3 n 2.1 - ettfest Xt) = a-bettet CS Came