Question

求微分方程a^2*w''+sin(2πx/L)cos(2πx/L)=0,w(0)=3,w(L)=6

求微分方程a^2*w''+sin(2πx/L)cos(2πx/L)=0,w(0)=3,w(L)=6求解微分方程a^2*w''+sin(2πx/L)cos(2πx/L)=0,w(0)=3,w(L)=6,其中a，L为常数

Answer 1

a^2.w'' + sin[(2π/l)x].cos[(2π/l)x] =0
a^2.w'' = -(1/2)sin[(4π/l)x]
w'' = -[1/(2a^2)]. sin[(4π/l)x]
w' =∫ -[1/(2a^2)]. sin[(4π/l)x] dx
= [l/(8πa^2)]. cos[(4π/l)x] +C1
w=∫ [ [l/(8πa^2)]. cos[(4π/l)x] +C1] dx
= [l^2/(32π^2.a^2)]. sin[(4π/l)x] +C1.x + C2
w(0)=3, =>C2 =3
w(l) =6
6=C1.l +3
C1 =3/l
ie
w=[l^2/(32π^2.a^2)]. sin[(4π/l)x] +(3/l)x +3