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ENGLISH VERSION OF QUESTION IS RIGHT BELOW MALAY VERSION.
I have done questions a and b. pls do check my answer. however, what I am not sure of is, how to find A and B to form the solution. i am stuck at the Fourier series part to find A and B. when I tried, I got A as 0. for B, I got a long function. do we need to give a precise value of A and B or do we need to just give a general solution in terms of n? pls help. thanks!
it is question c that's hard for me.
Soalan 4 Question 4 Diberikan suatu persamaan gelombang Given a wave equation a²ua²u əx² at²' dengan syarat-syarat sempadan with boundary conditions dan syarat-syarat awal and initial conditions 0≤x≤ 4, u(0, t) = 0, u(4,t) = 0, u(x,0) = f(x) = sin", (0, дир = g(x) = a, atle=0 (0, t> 0 t> 0 0≤x≤4 0≤x≤1 1≤x≤3 3 ≤x≤4

(a) Dengan menggunakan kaedah pemisahan pembolehubah, dapatkan dua persamaan perbezaan biasa (ODE). Nyatakan andaian yang dibuat dan gunakan sebagai pemalar pemisahan. Using separation of variables method, obtain two ordinary differential equations (ODES). State the assumption made and use -λ as the separation constant. (3 markah / marks) (b) Dengan mempertimbangkan tiga kes: λ = 0, λ = -α² <0 dan λ=a² > 0, selesaikan persamaan perbezaan tersebut. Kemudian pertimbangkan satu kes untuk mendapatkan nilai eigen dan fungsi eigen. By considering three cases: λ = 0, λ = -a² <0, and λ = a² > 0, solve the differential equations. Then, consider one case to obtain the eigenvalue and eigenfunction. (7 markah / marks) (c) Selesaikan masalah nilai sempadan dengan mempertimbangkan syarat-syarat awal dan menggunakan siri Fourier. Solve the boundary value problem by considering the initial conditions and using Fourier series. (5 markah / marks)

NO top klx 1002, Question 4, case TX" = T" X X X a=1 b=0 C=-d² X" X a=1 b=0 codz X" + 2x = 0 X" = -2 X D x" = -xx Case 1 Cwhere 2 = b) : T X" = 0 x' = 4x X = C₁ x + C₂ T 2 ( where 2 = -2²): x" + 2x = 0 x" - dex=0 X" +2x=0 x"+d²x=0 Case 3 Cwhere 2 = α²): m² +α² = G X = (5 sinh dx + 4 cosh dx PY Q 2019 v=XT= (₂x+₂) (C₂+ + 4) T" T m²=-2² m = ±di x = cq cos 2x + Gio sindr T" + 2T = 0 T" = -AT T" +AT=O T"=0 T' = 63 € T = ₂² +4 = -2 UAXTC CCs sinhax+ (6 cash ax) (G sinh at + cg cash at) 9=1 b=0 C=d² a=1 b=0 T" +2T=0 T" _d²T=0 c=-1² T" +2T=0 T" +d²T=0 DATE: To G sinhdt + Cg coshat m² +2²=b mẹ đại T= ₁1 cos dt + C₁₂ sin dt U = XT = (Cq los dx + (₁0 sindx) ((11 los dt + (₁2 sindt) POP bazic

Applying brindary conditions, 1 Uco, t) = 0 = Case 2: Case On CCC₂(+4) OF C₂ uc 4, t) be (C₁x + C₂ YC₂4+ C4) D = (46₁+ C₂ ) CC₂ + (4) 0 4C, +6₂ 0 = 46₁ +0 0 = C₁ - No meaningful solution. Ca2+8)) UCo,l) 20 (4 sinh dat coshaz)CG sinh dit coshae) 0= CC sinh dễ the coshat) 0 = 9 UCA, ĐE0CC, sinh axet co coshaw) CG sinh xecg coshat Cô sinh do cây sinh dfly cosh di cọ sinh de 0 = (5 sinh 4d 0 = 65 - No meaning ful solution 3: Ewhere uco,t) = 0 = (ca cos dx + (₁0 sin dx) C C₁1 209 dt + C₁₂ sin at) DF Ca C C₁1 Cos de + C₁₂ sinat) 0=Cg UCAA) = 0 = C10 Sin Ad C C₁₁ cos xt + C₁2 sindt) 0 = C10 sin 42 0 = 90 "No meaningful solution. But, we need a solution. Hence, C₁0 70. sin 42 = 0 AL=12 POP bazic eigenvalue

nex (411 y(x, t) = C₁p sin ny cos nat + C₁₂ Sin 90 9₁ cost + 91 hot + 910 (12 sim not t) sin ny 454) A los no t + B sin ny may +) sin not be DATE: ← eigenfunction