Question

(1 point) Find a particular solution for each differential equation y" + 3y = 5t, then yp = y" - y = 3et sin(t), then yp = y" + 4y = cos(2t), then yp =

Answer 1

diff.eq. find a particular sol. for each 9"+ 3y = 5t sol y+3y=5t - A.E.S m²+3m=o mcm+3)=0 m=o,3 Je = c +cest Now we find up by method of undetermined coefficients. Yp - (At+B)t -- 9p = Attet - Yp' = 2At+B p=2A put these values in diff.ea (2A) + 3(2At+B) = 5t 2A+At+33 =56 (ZA+3B)+GAŁ -51 cauating coefficients of similar teams on both sides, st coefficients :-) GA=5 =) A = 5 2 constants :-) 24+3B=0 215) +3B=0 5+3B-0 => 3B=-5 الا انا را اما :yp: 5+'+(-5)+

Now ① gluy= betsint , hou Up = ylly= 3e sint - 0 A.E. is m?i=0 m=1 ..m= I) = m = 1,-) You cieti cset we find Sp, by method of undetermined coefficients. You et Cacost+Bsint) yp = Aetcost +Betsint → yp = a[etcost-etsint] +B[et sint+etst] => Yp' - Aet cost-Aetsint + Betsint + Betcot .. Yp! = (A+B) (etcost) + CB-A)et sint Yp" = (+B) (etcost-etsint) +(6-4) (et sintretcst) - Aetcost+ Betcost - Aet Sint-Bet sint + Betsint-netsint+ Bet cost - Aetcost 2.4p"'- 2Betcost - zaetsint put these value in given deq. ① (2Betcost-2aetsint) - ( Aetcost+Betsint) = 3etsint (2B-A)etcost + (-2A-B) et sint szetsint eauating coefficients of similar teams on both sides, set cost :-) 2B-A = 0 2 et sint :-) - 2A-B = 3 2B-A -0 =) 2B-A =) A-2B

-2A-B=3 - 2(2B)-B=3 =) -53=3 =) B: - A=28=A = 2(0) = -6 - A = , B=-3 ->p = get cosct) - Be'sinct) -Yp - let-cascto a sinct))

~ y'44y = cos(at), yp: solo y+4y=coscat) - A.E.18 m²+4=0 m²=-4 m=zzi .: 9c-4,28(at)+cą Sin (et) Now we find 9p:-) by method of undetermined efficients. yp = [A cos(at)+ B Sincat]t Yp = Atcostat)+ Bt sin(at) ->$ Yo'= A(-2+ sin(at)+ cof(2+)] +B[ztcsczt) + sincet] ... Yp = -2Atsin(at)+ACOS(26) + 2Bt Cos(25)+B Sincat) Yp":-2A(2t c8c2t)+ Sincat)] + a[-25inc2t)] +2B [-2t sin (2+)+c83c2t)] + 2 B COS(26) =-4At c8(2t - 2 A Sin cat) - 2A Sin (at) – 4 3 t sincat) +gBc8(26) + 2B C8(26) 90!! = -4 Atcecat) - 4 A Sin (2+)-430 sin(at)+43 68(26) put these values in given d.eq. F-4 At c88c2t) – YASin (2+) -4 Bt sin(at) + 4B C8(2)7 +4 [At coBc2t) +Bt Sin (2+)] = c(at) 5) -4Acozt) – 4 A Sin (2+)-4 Bt sincet)+4 Bcbcat) +44t c8(2+)+48t sincet) - $(2+) =) -4A Sin (2+) +43 c8(26) = cos c2t)

cauating coefficients of similar teams on both sides s Sinat :-) -4A=0 ) A=0 2 coat :-) 4B=1 ) B= // A=0, B=4 Yp= 1 + Sin cat) :yp - sinczt)