Question

به الا وا/ ** را جی کے . (t) في * مل (ی) - X = 0 X(0) > 0 Sinat dx + ** dt 0 = (ه) و

Please use the LaPlace Transform Method to solve both equations

Differential Equations

Answer 1

40) y" #16y a } O , 22 eos (4-2) , >,2 -0 solution :- Now can be written an, y" +16 Y a 0.14-2) + cos (4-2) Ult-2) ,:. y" +169 coslt-2) U (t-2) Now, we state some of reflace s v L eostat} 32792 Then, e-2s 24]f(5) es} vi properties transformation OL}{"(t)} = 3² L{FH)} SFI0) -F'(o) © 23F'(4)} = 54]F(t)} - FIO) 4 FW!) Ult-2)) = 2-25 4/F (4+2)] vil If, 2-1/F13) = f(4) 0 - 1 aastat }, sincal) = F(t-2) Ult-2) 6 2 1 3 tan } a cos cat) Now, we faking kaplace transformation of , we get 4}*" +1}= 4 cosce-2 Uc+-2] L}%"}+16L3732 e-25 L eos (+-2+2)} 5228964)]-54(0) –Y'60) +162393 (3+16)2393 -6,5-221 (54+16) 4303 e-2s e-23 LJ east} [het Let, Y (0) =9, 5²71 y'(0) - 02 s e-2s ] S teis tez 3²+1 S te2.5716" ei S e-2s s716 2343 2 (5+1) (5'+16)

2S S 4 제6 S 15 +2 5416 34 S S e2s 49 L593 e23 15 5+ 546 15. S+ +2 3416 헤 (3) NOW, S LI +1 3 cost 도 ~ e-24 13 S 제 cos (2) ut-2) 으 5 네 2 Cos4k He } / S-5 / - <s44 ~ e-25 COS (2) a ut-2) →⑤ 1 N es NOW , 넥 LI 비밟 sinut I sin 3 5 192 (2) ~

nous, Talking Inverse hofleve trago formation of @jue yed les erzs] ~ (*) = LI e-25 } 15 h S 32 +16 S +4, L 13²416 } + 92 242 37416 13 2 ts eos 4-2) U(+-2) - 15 eos 464-2) Ult-2) et sinh [cusing 9,6 6] te, eBut + si y(t) = 75 [ 003C7-2 eos 464-2)] U14-2) +2, 28344+ mesingt .: The solution of the given differential equation ois, + Y+) = 15 [cos (4-2) – 285(44-8) ]ult-2) + 4,283(44) + = sin(44), where, C, = y(o) f C2 = yllo) [sallation ] 5 -X =0 d- dx 370 X 10) = 0, y(0) ty= sin at Salution :- dy 2 x=0 dx de tY = sinat ③ Differondeiating ② word t, we get d22 dt² 2 2 eos at te - 6 [using d22 de2 tn 2 2003 21

Now, Here, 260) >0, y lu) Then from ② , 2'60) +7(0) = sinco) 2010) 2 0-0 Thus :X'10) 20 , we get, x(0)=0,&'(o)=0 transformation of B, we get d2x ta 2 d t2 Now, taking dafluee L} +] 4/200326] 21*"] +21332 223 403 (24) 524323-5 5.2(0) - 410) +4390)= S 2 32+4 - 2. s2+4 1571)4393 - 5.0-0 L303 z 2 s (3244) (5+1) 3 L303 25 3 2 [ (52+4)/571 1] :] 1 23232 2335 [ 5+1 s²7 --> 5 Now, taking inverse da place s Lol L323 2 2 3 3 3 3 3 3 3 3 3 38444 transformation of, wegu XH) = 2 2 2 2-3 384} - 23 23 203 (26) + 4 sin (24) A s² +22 ult) = 23 CO3 + Now, g'(t) = 35 sind

Now, from @ 2 Y(+)2 a'(+) + sinet 2 singl) - sin let ) + sin (27) 2 w/n 8 94) 2 2/3 sin(t) + sin(24) :. The solution of the given differential equation is, X(t)= cos(d) – 103 (2+) H(+)= sinct) - sin (24) [ans co/po wola