Question Number 100891 by bemath last updated on 29/Jun/20

u_(tt)  = u_(xx)  − 6x ; 0≤x<π , t>0  u_((0,t))  = 0 ; u_((π,t))  = π^3 +3π  u_((x,0))  = x^3 +3x+3sin x  u_t (x,0) = 0

Answered by bramlex last updated on 29/Jun/20

u_(xx) −u_(tt)  = 6x ⇒(D_x ^2 −D_t ^2 )u = 6x  u = (1/D_x ^2 )(1−(D_t ^2 /D_x ^2 ))(6x)  →Taylor series (1−x)^(−1) =1+x+x^2 +x^3 +...  u = (1/D_x ^2 )(1+(D_t ^2 /D_x ^2 )+...)(6x)  u= (1/D_x ^2 )(6x+0+...)   u = x^3 +g_1 (t)x+g_2 (t)  u(0,t) & u(π,t) → g_1 (t)=3 , g_2 (t)=0  u_1 (x,t) = x^3 +3x   u_(xx) −u_(tt) = 0 , set u=∅(x+at)  ∅′′(x+at)×a^2 =0→1−a^2 =0  a = ±1 → u_2 (x,t)=∅(x+t)+∅(x−t)  u=u_1 +u_2  ∵ u(x,t)=∅(x+t)+∅(x−t)+x^3 +3x  so u(x,0)= x^3 +3x+3sin x  2∅(x)= 3sin x → ∅(x±t) = (3/2)sin (x±t)  u(x,t) = (3/2)sin (x+t)+(3/2)sin (x−t)+ x^3 +3x ★