Question

QUESTION 3 After use Laplace Transform to transform the following initial value problem X" +x=e-t, x(0) = 1,x'(0) = 1, S-2 you should get X(s)= (write fraction as (S-2)/(5-4)(8+6) for -). Then, find (s-4)(8+6) x(t)= L-?{X(s)}= (write 5/6 by 5 -30 6' e^{-3t} by e and sin(2t) or cos(3t) by sin(2t) or cos(3t)).

Answer 1

civen that :- x'+xzet x (o)-1, 2 x (0) 21 > { Now: x'+x: et Apply Laplace Translam. kfz"+r} = a[et] 2{2"} + hhx} = 2{et 5*L[7403 -5 260) -2'60)] + 1245)) 5+265) - 5-1.+265) = ) sti sil 1 st + 5+1 3+1 (5²+1) 26) - (84) a(s) : i +32+5+5+1 5+1 ७२ +2.5+2 = A st ais) = $?+25+2 (5+)(8+D using Partial fractions Seite (5+1) C52+1) A Cs+1) + B5 (s+i)+(5+1) Compaings": ) Componing is :- | Companing A+B 2 Bic=2 Atc=2 2A = 1 2 + A+c= 2 A=/ c=2-Y2 B +$ 2 B=₂ A=B 2 s +25+2= z +6=2 6 = 3/2 A-B=0

A BSc t(s) a + Sti 31 Y S + 3% a(s) - + 5+1 52+1 + S 5²41 + 3 2 2 5) (s): 1 Šti Apply Inverse kaplace transkim 4."(26) = 4" [1 + 3+ 4+1 2 3}+-"{1*] t alt) - Žét + Cost + Cost + 2 sint ALE)=z 2. (et + lost + lost #3sint) (ar) .t Si alt)2 Ž é to cost + sint. 1