Question

6. (a) Determine the inverse Laplace transform of F(8) 2s - 1 32 - 4s +6 (10 marks/ (b) Solve the initial value problem using the method of Laplace transform. 7+10y = 0, y(0) = 0, (0) --3. (20 marks) (c) Solve the initial value problem: 1 dy dy 4 dx2 + 4y = 0, y(0) -- da + (0) = -1 120 marks)

Answer 1

Invene Laplace transform Fes) 2S-1 5²-4546 s² 4546 s²-43+ 4+2 (5-2) 2+2 25-1 = 2 (5-2) + 3 Fes) Take I'lfesty 205-2) + 3 (5-2) 2+2 nurse laplace 12 265-2)+31 1 (6-2)²+2 Inresse skipting them s-2 S at il as + 3 s2+2 24 25 ti 412 sta? s2+2 fet) 2 cosszt + 3 sinset

(yet) 2-3 & sin hat Initial value Problem Y"-ty't loy y col=0 y (o)=-3 y"-7ylt loy Take Laplace transform {y } -72{y'} +10Lhy}=0 Sy(s) - SY (0) -t'co) - 7 [ST (6) -760) tlores) = 0 (5² 75+10) 7 (s) +3=0 YIS) = (5-7 Innesse Laplace [/27 (3} 12 -)? g2-9/4 #12 (-3) sinhat 3 s²_75tlo 3 s²-75+49 .49 +10 3 Now 768) = 2 4 9/4 7/2 z x

y"+y' +4y=0 yeo) = -1 4 y'co)=-1 Take 2 =0 - y" + +y'+4y=0 Laplace transform L}%" } +42}y'} +4hz4} s}(s)-stol-yl) + 4 [5725) -760)] + 4715) = 0 (5²+45+4)*(8) + s+1+4 () = Yes) (s+45+4)= -2 -2 -3 Yes I -5% - 3 [+45t4 (5+2)2 :(5+2)4 res) = Take -- (5+2)-2 (5+2) 2 Invesse Laplace transform (5+2)-2 (5+2)2 I press} - L 1 ' ? ezt Il} -2 -2 } eti 2 - } 2S 52 -2t -2t -e-zeit -2t [ 3-2] 2