Question

5. Use Green's function to solve -(") cos(), y(0)y(1) 0

Answer 1

given boundary value poro blom 9n addi Paga boundasy conditions ylo)=y0)-0-12) Og Ca (2) with the Satisfy Considesr the associatedd boundary value paroblem -1"-4-0 C12 G( biGa I1 091 (02+1)Y=0 Sbject to the boundasy Conditons Cii) The Y(o)=0 4/a) Y() =O We shall orst ind the Goreen's Sunchon os the above menti oned boundasy value problem 9iven by (3), 4la) & 4(b) The auxiliasy ea,uation of (3) is m2+ l=0 gonesral salution of (3) is whesr o9 d Hen ce the (5) AC[) + B Sin (oc) (s) u and Using B.C4(a) Putting get Pattng 9et Fom (6) and (7), A B0. Hence (5) yields the toiuial soluton yla) 0. Tharstse Ciny G -- (6) A=0 OC in (5) and USmg B C 4(b) (7) Co Cwe and lot ny Graen's fmction exists fasr tha boumdary value poro blbm 'given by 13), 4la) amd 4lb) and it is giuenby 91Ca420 +92 Sin c Oc t Glxst)

9n addition to the above poropesty (e), the pro posed Satisfy Grean 's Sunchor mustals o the dollowing thoree proposches CiD GOC,t) is Continuous at oc= t io biCas t + b2 Sint = a,st +a2 Sint lb-911ast +lb2-92)Sint -o Ci) The desrivahiuo os G has a discontinuity o magnituda-polt) whesre poloc)- Coosficiant os the hig has t Ondesr cerViatue in (3)= ,that is G at the p0nt x=t t-o -b,Sint +b2 ast -(-a,Sint+92Co t)- lb1-a) SInt + (b2-92) Cost-1lo Cim) G(a,t) must satissy the boundasy Conditions yla) and 4(b)that tS Cif) So that Gl0it) O So that biCayb2Sinl-o -12) and Git) =0 and b2-92 C2 13) -(4) let b-a=C - C, Sint +C2 Cast =0-(15) (14)& 1I5) by Coross mulhhliahon Page- C Cas t + C2 Sint to = O Salving we have method

Then we knownthat the Solution of the given boundaoy value poroblon c)-a) Y(oc) Glxt) plt)dt () where plx) = y So that lt)= G t) Hence the sreguised Solution is given by Page-y Yla) G(xt) C t dt Saa,thet ldt Gt)Gaytdt+ Gyt(Cn 2c-tanl Sine) Sint dt Cot (Cast+tanl Sint)Sine dt x)S Cost Sintdt -(Con 2c-taniSinx o Sinoclf Ca2t dt + 1 tan1 Sint otdt + RSint tdt (a4x-tani Sin c) 3) 2

P= $70, t= 10 y2ans, n = 6/0 119 +Sinc p2afPtdt Sino tan) 2SintGtdt 2 2 (Cospc-tanl Sinec P 4241S 2- +Sinc rut G12t)dt +Sinzctanl Sin2tdt 2 2 OC Cafpc-tanl Sinc + Sin 2 2 2 O Sinoc tanl 2 ( (g Coy c-tanl Sin Co120+ 2- SC uS+ 2. Coy 2 Co 2 +Sinzc tan) 2 Ans 2 2 Page-S

is an iode (h) dantty Eauating to 3aro tha C2 74is -Cost 1lence C=Sint C201Cyt th b-91 Sint by (13) b2-92=-Cot by(13) 17) Solving (1), 12), C16)& (17) we have biSinl-b2 Caj 1= o Sint sinlb2Cy O bg Cas Sint Sinl b20 Sint Sinl bz tanl Sin - tanl Sivt-a2= Coy t -92 Cas t tani Sint tan! Cost a2 the above valuos in (6) Substitutng cwehave - tanl Sint) Cas o zt Gt)(Gt (Sint arc tanl Sint Sinoc tx1 Sat- tani Sint) Gasx o t (81)- Sinz) Sint tecs Uanx + tanl