Question

(16 points) Consider the equation for the charge on a capacitor in an LRC circuit + 9 + 169 = E which is linear with constant coefficients. First we will work on solving the corresponding homogeneous equation. Divide through the equation by the coefficient on and find the auxiliary equation (using m as your variable) =0 which has roots The solutions of the homogeneous equation are Now we are ready to solve the nonhomogeneous equation +184 +809 = SE. We will use e-10 to do roduction of order (it doesn't matter which one of the homogenous solutions we choose), which will give us a particular solution of the nonhomogeneous equation. With this choice the particular solution has the form y, Me-10 Then (using the prime notation for the derivatives) So, plugging , into the left side of the differential equation, and reducing, we get +187, +80y, - Using this reduced form, moving the term e-10 to the right side of the equation, we can rewrite y' + 18y, + 80y, = 5E in terms of us" and was At this point to solve for u we want to integrate both sides, but to do that we need to integrate E. We will look at two special cases for E so that you get the idea how to proceed.

Answer 1

Giver dq 18 dq +169 - dt? 5 di + The corresponding homo egn is Id²q + 18 dq +1692 5 df² 5 dl throughout by 1 3 we get dq + 18 dq. † 809 de² dt Its auxiliary eqn is m’t 18m +80 = 0 Now solving m +1&mt 80 mit löm +8 m+80 = mim+10)+ $ (mto)=0 → (M+8)(m+10)=0 =|ma OY m -10 axe the roots. -86 -106 land we write The sol. of homo egn are ~lot + Ge d²q +18 dq + 809 = 5 Now solve the non-homo . egn: let us di-2 de -lot ue Put Yp = Then yp' -lot ue -lot loue -lot -lot -lot -ot + looue loue" t-louet u" of diff egn. we get Plugging this to the left hand side or 11 - 10 sue coule of loul :-(06 e gp" +18 yp' + 80 yp tlooué tigué ajoues -lot tue so 11-Ot-lo zou'e ue -lot #love u + laule -lor 18oue troue -105 u" e lot aué tot

-lot Y;" + 184p't soup u'elo Pule --lor 9p' +18 yp' + 80yp - (u"-2u') & e Yp"'+184p et 80 up 5 € will become -lot (u'-24') e lot u-zu: 15€ Casel when e a constant lot u"-24' = 55e with respect Integrating both sides 10 101 56e Ee ü'-24 = be 10 This Linear diffego a ý of the form The standard di linear diff egn factor dy with integrating da SPcuddu Spoxdx SP code and sol ye dx t. Constant + P(y = Q(x) e Socks Sponde linear diff egn u with our standard Now comparing our form we get PCE) = -2 So integrating factor e s-adt .at e u we get Solving for Tot -zt e e dit b ( constant here 6 • geven ue

-at st ue E e dt + b 8t -at ue + b 2 8 tat 86 727 Ee ..e 16 t be 26 U E 16 10t et be ~10아 become ue will Ip -106 lot 26 t be e se) e 16 Up -SE + be alm + be 56 1645 16 Y 5€ 80 BE be + Case a when sin (PL) CAC voltage) lot Then a "zul iot 5Ee 5 Sin pre we get Integrating both sides -24 = 5 5snpt iot e de 5 sinprelor de ax NO TE I e + Constant We will s ea'sinbude asinba-brosbx a²46² prove this later. we have to evaluate Seelofanpt du.. so. on Companson get a = 10 p b=p. Here SO we

lot u-au 10 Kippt -- prospt ]e 100+pa PCE) = -2 which u again linear diff. eqn with Splidt s-adt factor e -at e e so Integrating -at Integrating factor is lot and Qit) - 150 sinpt - 5pcospt) e 100+p?. Hince sol (so finpt - 5poospt)cot et alt + b 100+p2 -27 u e st est)di]+ Sprospreat 21 ue [ (sokiopte 100tp? b + -at u 50 J Simpten 5p [ cospre oralt 100+pa NOTE 2 ar Constant acosbx + b sinbx + Jean Cosbx dx q² +6² be provided later) note O we using jesinptat. (Proof will get 8 Sinpt - prospt 64+P2 st ipt Jest of using note we get 8 cospt + p sinpt 64 +p2 SK Je rasptate (8.cospt + poopt

(400 simpt- sopiospt - 4 opcospf-5psinpt) + be'st so 0 becomes -26 ue painpela est frospttps:opt to 64+P²1 lobip simpt - pcospe] -58 *pal low e & beet w 400 sinpr - 5oprospt - 4 Oprospt - 5p simpt 64.4p² 100+p2 do -100 ue will become Yp 자 OL e ( 400 finpt - Soprospt- 4opcospt - 5pšinpt) +! (100+P) (64483) (100+P3)(644PM) -st པུ་ཁ་ ཊར་༨༨༩) fo sinpt - Topcos (pt)=&pcospt -pšinpt)+be (100+)664773)

NOTE 1 Proof I - seats inbx dx & Sinbxem - f bcosbx de de ax १४ Sinbreax b cosbx.e-s-b sinbx.ea da a a 2 Sinbreakdx Sinbre al b e cosbx a? (a sinbx - bcosbx) ax b3 Il Since I = Se "Finbadx) q² a² an I + b² I a sinby- brosba a² :) e ean lasinbx - bcosbx I (it lan) 2 sbx] - boosbx) an e I a siriba-boosba a? lasinbx-bcosbx a²+62 eax Note 2 Proof I, Jean cos bx dx cosbx.com - [beinbrechen di a an 9x cosbext es sinbxearede cosbxe Sinbu-en-f bcosbx de Cosbe en het besipbxecuve scosbxeady + a [Sirbnie a

acosbet bsinba be I, a² an a² (aco bsinbx Jeax » (I, + bay) = (acosbx4bsinbox soba) (te an е T. => acosbat bsinba' a² acosba+bsinbre a²46² * ) е.
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