Question

use the Laplace transform to solve the given initial value problem:

Questions 8 and 12

8. Y" – 2y' + 2y = cos t; y(0) = 1, y'(0) = 0 9. y" – 2y' + 2y = e-t; y(0) = 0, y'(0) = 1 »Answer Solution 10. y" + 2y' + y = 18e -*; y(0) = 7, y'(0) =- 2 11. y(4) – 4y" + 6y" – 4y' + y = 0; y(0) = 0, y'(0) = 1, y"(0) = 0, y" Answer Solution 12. y(4) – y = 0; y(0) = 1, y'(0) = 0, y" (0) = 1, y''(0) = 0

Answer 1

s 3+1 SA 6 aiven initial-value problem (INA) is y"_24'+2y = cost i yo)=1, glosão Applying Laplace pransform, then Lly"-24'+2y} = L{cost! => Lly")_2215") +2L1%} = 28 =) ? YLs)-sylogyl)-2[syks) - Y103] +2415) - => s² y(s)-5-0-25YLS) +2 +212S] =. -> (5-25+2)/(S)-5+2= -> 184-23+2) YS) = a +5=2 = 5+82-254+5-2 => (3²_25+2) YLS) = 33-25²+25-2 s²4 -> 415) = 3_25² +25-2. (52+1) (3²_25+2) partial Fractions 5325²+25=2 AstB CS+D (5²+1) (3²_25+2) 8241 8².25 +2 ²+1 + -> 3_252+25-2 =(As+B)[-25+2)+(cs+o) [8S+1). -> 93-252+25-2 = late]53+(-2A+670)5? + [24-28+c]s +(2340) compare coefficients on both the sides, then Atc=1 -2A+B+D-2 2A-2B+c=2 23 +2 from şi col-A use this in o. 2A-28 +1-A=2 => A-23=1

From ) D= 2-2B use this in. -2 A+B+ 2-23 5 - 2 => -2 A-CC ~ 나 CD 2AA4 selving and, 2A-4B=2 2AA B - -SB=-2%B= As As 며 내hn A응, 여을 A= 9 then c= -4/5 substitute A=9, B=235 , 2=-4 and D= 6 in 224252 월마하2Ste) 웨 SR2S42 Then 4S+6 능 구르 + USE 9. in D. Then 늘 6 2-25 y=윭 9 굵 Applying Inverse Laplace Transform. Then 느 ) 빠 22S42 대응훼를 훼봉 윭 == 대체를 대체놓대떎 리에 => 4t) = q cost+ 2 sint- >> %) = cost + sind ecl) + l => YLt) = as cost + 2 sint-Hs et cost +2 zet sint. (1

.:: ytt) = ŚLacost 72 sint-4et cost +2 etsint). Answer 414} = 'Ś la cost+2 sint-4etcost +Retsint). (12 civen initial-value problem (IVP) yL4)_y=0; y,0)=1, 920)=0, 4(0)=1, yll(o)=0. Applying Laplace Transform, then L1y14)_y} = Llob - L1yL4}}-L1yb=0 => styls) - 33 yol-5² y/o)-sylloj llos - Ys] =0 => (34_1) YLS)-330-s-0=0 -> (54-15 YLs) - sts -> (3341) (5-1)(S+] Y [s] = sisty -> (5-1)(3+1JYLs) = 5 => Yls) = (5-1) (5+1) Appleging partial fractions B SI (S-1)(3+1) => s= ALS+1) + B(S-1) If salithen 1=2A => AE Ź If so-1, then-1=-28=) B = Ž substitute a-Ź , B=/2 in. Then

5 + s (5-1)(3+1) using in Then Y(s) = Ž ša tž si Applying Inverse Laplace Transfoom. Then y(t) = ['l È + Ž 4 ) Syd)= {[c15]+[15] ->yt) = Źlettet). yet) = cosht Required solution. Answes Yt) = cosht