Find the solution to the initial boundary value wave problem
∂2u∂t2(x,t)=4∂2u∂x2(x,t),\displaystyle \frac{{\partial^{{2}}{u}}}{{\partial{t}^{{2}}}}{\left({x},{t}\right)}={4}\frac{{\partial^{{2}}{u}}}{{\partial{x}^{{2}}}}{\left({x},{t}\right)},∂t2∂2u(x,t)=4∂x2∂2u(x,t), 0<x<π, t>0\displaystyle {0}<{x}<\pi,\ {t}>{0}0<x<π, t>0
u(0,t)=u(π,t)=0,\displaystyle {u}{\left({0},{t}\right)}={u}{\left(\pi,{t}\right)}={0},u(0,t)=u(π,t)=0, t>0\displaystyle {t}>{0}t>0
u(x,0)=2sin(3x)−3sin(7x),\displaystyle {u}{\left({x},{0}\right)}={2}{\sin{{\left({3}{x}\right)}}}-{3}{\sin{{\left({7}{x}\right)}}},u(x,0)=2sin(3x)−3sin(7x), ∂u∂t(x,0)=−4sin(6x)+sin(10x)\displaystyle \frac{{\partial{u}}}{{\partial{t}}}{\left({x},{0}\right)}=-{4}{\sin{{\left({6}{x}\right)}}}+{\sin{{\left({10}{x}\right)}}}∂t∂u(x,0)=−4sin(6x)+sin(10x) 0≤x≤π\displaystyle {0}\le{x}\le\pi0≤x≤π
u(x,t)=\displaystyle {u}{\left({x},{t}\right)}=u(x,t)= Preview Question 1
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