Question

Determine & l{F}. F(s) = - 352 - 10s +9 (s + 5)2(s + 1) Click here to view the table of Laplace transforms. Click here to view the table of properties of Laplace transforms. £-1{F} =

Answer 1

First I have find the Partial fraction of F(s) and then easily find the required inverse Laplace transform of F(s)

I have given detailed solution.
solution'! To solve L {F} Where F(S) = -35²_los ta (3+5)2 (5+1) we will First fraction of finel the partial Fis) Now 2 -35-los 9 А 33 C 2 (5+5)*(5+1) (5+5) (5+5) (st) -35²-los +9 = A (5+5) (5+1) + B (5+1) + ces+5)? -33²-los+9= A(8+65+5) + B(5+1) + c c(st105+25) Now comparring Coefficients s², S and Constant terms from both sides respectively then we get +c= – 3 6A+Btloc= -lo & 5A + B + 25c = 9 Now will solve these equations for A, B & C.

From equation 4) pasa = 6A+Bt loc = - 10 6 (A+C) + B + 4C = 6(-3) + B+ 40 = - 10 { • From equ(3) Atc=-3 =) B+ 40 = -10 +18 =8 B + 40 = 8 anel From equation 5 A + B +25C = 9 5 (A+C) + B + 20 C = 9 =) 5(-3) + B+ 200 = 9 = ) B+200 =9+15 = 24 Bt 20c=24 B= 2 4 -200 substituting equation equation (c) into B + 40 = 8 =) 24 - 200 + 40 = 8 - 160 - - 16 =) (= 1 B + 40 = 8 = ) B + 4 = 8 = > B = 4

page3 A+C = - 3 => A=-3-c = ~3-1 Now substituting value in to equatron (2) A, B and c thea we get -35²-losta (5+ 5² (5+1) !! + 4 (5+5) S+5) 2 = F(S) (541) Now L-1 {F} = L'[ (5+5) 2 S+5) (5+1) L'(F) L ' (575) + 4 L'{ (5+5)2} 5t PI(F) == 42 +4 te -5t L- ' { šti? tët Rude L'Estalat and F(t)= L-'(F) 4 & 5t -5t L-{{staj/= teat +4tet it te OR # =) L-(F)= 4(t-1)-5t tet This is required solation!