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find u(x,t) corresponding to the initial values f(x)=sinx,g(x)=0,when u(0,t)=0 du/dx(pi/2,t)=0,c=1
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find u(x,t) corresponding to the initial values f(x) = sin x, g(x)=0, when u(0,t)=0, (αu/αx) (π/2),t)=0, c=1 the wave equation u=1/2(f(x+ct)+f(x-ct)) and since g(x)=0 so we don't need to consider it.
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\[U=1/2f((x+ct)+(x-ct))+c/2\int\limits_{x-ct}^{x+ct}g(x)dx\]
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sure
Confirm: the wave equation is \[U=\dfrac{ f(x+ct) + f(x-ct)}{2} \] or \(U =\dfrac{1}{2(f(x+ct)+f(x-ct))}\) One more thing: \(U'_x(\pi/2,t) =0\) and \(c =1\) Question: \(c=1\) is for \(U'_x\) only or for the original U(x,t)?
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