Question

(10 points) This problem is related to Problems 8.16-21 in the text. Consider the differential equation y" (t) + 16y'(t) + 68y(t) = –20e-4t u(t), with initial conditions y(0) = -3, and y'(0) = 4. Find the Laplace transform of the solution Y(s). Write the solution as a single fraction in s Y(s) = help (formulas) Find the partial fraction decomposition of Y(s). Enter all factors as first order terms in s, that is, all terms should be of the form on, where c is a constant and the root p is a constant. Both c and p may be complex. Sep Y(s) = Find the inverse Laplace transform of Y(s). The solution must consist of all real terms. (Remember to use u(t).) help y(t) = --! {Y(s)} = (formulas)

Answer 1

Sod y (t) + 16 y (t) + 68 g (t) = -90 e 4 it u(t) 960)=-3, I'co)=4 take daplace transform. 8AYCS) - SYCO) - Y'(0) +16 SYCS) – 16 Y ( +684) (5+4) = Y(s) (32 +165 +68) = - 30 - 35-48+4 (S+4) Y(s)(824168+68)= 30 -35-44 (S+4) Y(S) = - 20 -35-49 (5+4) (8² +168+68) (82 +165 +68 - 35-44 (S +4)[(+82+(212) (+85°ta} -! + + (8+8) 'S+4) (S+)2+(2) 2 S + 8)2+(27 - 3 (S+8) (5+82 +212 (5+8) 24212

YOS) - -1 -18 +(-2) (5+8) (3+4) (S18) 2+ (2) 2 (848)2 + (2) 2 inverse laplace tape y(t) = - e-tet - a 6-8 it sinact-2 0-8 it cosant