Question

Apply separation of variables and solve the following boundary value problem 0 < x < t> 0 t>O Ytt(x, t) = 25 yxx(x, t) ya(0,t) = y2(7,t) = y(x,0) = f(x) yt(x,0) = g(x) 0 << 0 <r<a

Answer 1

This is a standard form of one dimensional wave equation. We solve it with the help of seperation of variables and we write down the solution of the problem completely and doing this we also used method of solving a ODE.

SH ito Рест, ST Y(x,0) = fox) } o cxcT = Ein] Nato each The problem is Yue (xt) = 25 Yax (x,t) Boundary conditions ane restron RO Dr. J (0,t) = x (TT, t) = 0 20 Wien Initial conditions 0 are, 7t (2,0) = g(x) FC 360) doo solution: ht Assume that a solution of ① is of the form Jee,t) = X(X) TIH) [ Sepenation of vamables] Then from we get, x(x) "C) = 25 x'x) TT) Code ) : * = 5 as 2,t are indepewent ramables xo it est leste possible only when each side is equal to a (day). we get two KX = 0 0-300 dhe ? d² T' From a constant Then ODES: K d²x sedia a VI dit 2-25k T =0 ootostudio px (1) when ko (=p2, say) then x=ce & Gë --þx Nonts T = zespt & C e spt erot ott Keo Carp? say) then od x & caspx + & sinpx (1) when k=0 then (1) when ( T = G cos spt + Go Sin cft stilt = x = Ca x + Go T = 4,4+4216492 - [ By ODE solving ]

Now if we by I we Foro will an when where xo and Goitibros with apply boundary conditions given by get only non trivial solution (which is also consistent with the physical the sical nature of the solution) k<0 (= -þ?), le XXIXA YOX 4) = x (x)T +)); - ( Caspx + 6 simpx) (cy cos ipt +Ganspt) worot ont to Nostulace s tant amuaga lo we get, 72 (0,4) = 4x (94) =0 X = ( [" tap sa I went (-p Cz sinpx + p & Cospx) (t) | An Br Now applying A G C 20 -O Yt 66 CTIL) =O X [oo Tt) to fors ose Goldimar Com non triviality] dro Again, Z (2,4)=0 gives tong wala aldicool no out top IN WIT = 0 Loo & -o] -p ce simpre Telas 0 TS ; - pep Sinpr Tt) =0 we must have g to for non-trivial solution, also Tt)0 for non- trivial solution simport out 0 use sof) o Neto (11) 11, 12, Pomis 360 þ= 11,12,- (bto) WOOD 8. Hx, t) = Cocos (c, cos spt + Ce sin 5 pt) 6 %. ÞEntri nest pa = 21, +2.

we For whole set of solutions oo ar we will get yox, t) Jex, t) = [An cos($pt)+ Ben sein (opt) ]'cospa) u T where An, Bn are wefficients which as given by An So fox) bor ( px) de VIT Soth g(x) cos(px) de رم Br = 5þ TT MONT =1 [oo Applying intial conditious we get, yx,0) = f(x) Yt (x,0) = glx) I gives E An cospx = f(x) An = 2 r so f(x) Cos pre dhe Y+ (2,+) = È [ 55 An asin 5ft) + 5p Bn Co(6 pt) ] cosþz : Yt (x0) = 5p Bn Cospx = 9(2) D= þ=1 T & Bn 2 5рт So g(x) en pre de ] together give the regu required ined solution. Nel ad 180