Question

5. Use Laplace transforms to solve the following value problem. (10pts) (a) y" - y' = e' cost y(0)=1 y'o =-1

Answer 1

5. La Here the De y-y = e cost y (0) = 1, Y (o)=-1 a & {y' } & lest y'it) at 22 [est was to - Jesset yay at - Tē©ycey - 1093 y (t) at 2.0 = -st &S T(s) Se y(t)dt = 7 (8) 0 SY(S) - 1 212 - g) since I (o) at - St "(t) at 0 Testyles for was die [o - y los} + { let yet s sey Add Is y (0) - y'co) = 5246-54 my! (0) = 2 COPO Sys) 29 war smeely Coat Now , ya e cost - Gy(0)=d, y'(o)=nd NOW, abse cust} = seste cost at (e 0 I Set(sal) cost at O

-(S-ilt +(S-1) S6 (smut Sint OSO Sint at T2 -soigt cost -(S-1)t costat e Since el 2 (0-0) - (3-1) pē = (s-41-8 +1] - (-131 1 + (s-y1 = (S-1) (5.1) (S-1922 COS O Simo io Co Now hornsformation, Side saplace Use ing both diş y" - y'} = 0 { e cost? ho{y } - d{y} = hos et cost} s²y Cs) -sti -sr(s) + 1 = (S-1) s(S-1) Y(s): (S-) Is-2 + (S-2) Y(s) a (srl) s (S-1) (5-vg?+1) s(S-1) (S-2) s((S-1) +) S (S-1) "If we we Immerse taplace dansformation, Sonnon. Now set the

Now , ང་ ། ( ད ༤༥༡) (༩༤ - ) བྱ S ༡ s(༩s- ད༥) (s） བན ད -+ : sf - ད 2 sl cong 1) ༩ gང།) ། ། ། ན 2 (s)) 2 ༡ (༢-) + །) (s - ད༡ -༥༣) s(༤) པར (༡)： ༧ (༩- ཚེ ༡ ：《 ༡ ༣ ད) ༼ ༧ ༤༩༣ -os༌ ༩༥ Since ། at {fest} = f(t) then {f (s-a)} e F(t) ༡ ༽ here དུ Sint s24། S t -e simt ) $ Again (༥) e Cost - ༡༧༢་༩ Asain, 3 (<- 2) ༤ ༢ ༢S- གྲྭ - ཚུ་ S (5-1) A s-1

(5) - 6' (5) t 3 2-e at since (+3) = 1 and 2 (= Now, useing Inverse taplace transformasion both side of S-2 S (S-1) & Y S it (r(s)) = { ES (15-13'40) y (H = 2 (sint-east) + - + here eta (sint neast) + 2 - e t 11 So, y(t) = 2 (simt –cost) + this og Requined solution. 2