Question

d yt) 2 dy() +7- + 10y() u(t) dt dt2 y(0) 0, y'(0) = 3

Solve the following differential equation using variation of parameters.

Answer 1

The given difforental -egulion Solution (f)b = ROtBp t , y-3 dt dt2 2니 D&H0g= 4(8) cly-DR D+10)= 4f) (1) 2 auriliary eguation ot eguation (1) is uen by The D+ D+ 10 O Put pem mtRm+103O Ht5m+2m+10 z0 mim+ 5) + 2 (m+5)=0 (m+2) M+5) =0 Ond m+2 0 3II2 ghd-5 - 2 Hence the yoors are Then Complementary Junction af egurtion 1) 1s given by -s4 2 Ce (2) 24 B7om aboue eguahon (2), let Y,() = e Y2(P) S*ne soluytion of eauation 2). qnd To Sind the Constant 9nd C 4(0)00 y03 Griven that Put o n eg4aton 2) 4l0) + C2 OMCI+ C2 -(3) wiTto on both &ide n egyathin2 -SA (4 differutiate yA)-2G2-s3e Yl01-2-S Ce and Put &o

3 -201- S C3 57 201+503 --3- BTom equeHon (3) andl (), we 9et 20, + 2C2 O 20 +592 = -3 - 3 C 3 C2=-1 Purtting the val4e of c in-eguotion (3), we 90t C+ C2=0 CI2-C2 CI1 Pu-ting the Value of Ci gnd C2 n eqa tion 2, we 28 9ut which is 1egAired comtlementauy femctron. To Sind the patculor solution whare u= F1Om eay9Hon (1) h=H)f (-3)27 24 3 1 -24 SesA -Set2e whare utt ) is Uhkhouh

(f (FA P1Om eguatron (1 24 ult) e dt R)4() 31274 where u() is Uhknown

Putingr the value f u) 9nd u() in equation ), get 2ud 3 (dP SA R) dt 1 ( 3

Then the g eneal 8olytion is 414) 40(4) +Ypl) -54 -24(24uldt -54 3 P Angwey