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A vibrating string fixed at $ x=0 $ and $ x=L $ undergoes oscillations described by the wave equation $ \frac{\partial^2u}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2u}{\partial t^2} $ where $ u(x,t) $ represents the displacement from equilibrium of the string. Initially, the profile of the string is $ u(x,0) = \sin(\frac{4 \pi}{L}x) $ and its initial velocity is $ u_t(x,0)=8 \pi \sin( \frac{4 \pi}{L} x ) $.

My attempt at a solution:

Take $ u(x,t) = [ A \cos(kx) + B \sin(kx) ][ C \cos (kct) + D \sin(kct) ] $

$ u(x,0) = [A \cos(kx) + B \sin(kx)] \times C = \sin( \frac{4 \pi}{L} x) $.

So, $ A = 0; k = \frac{4 \pi}{L} $

Then, $ B \sin({ \frac{4 \pi}{L} x})[ C \cos (\frac{4 \pi}{L} ct) + D \sin( \frac{4 \pi}{L} ct )] $

$ u_t(x,t) = B \sin( \frac{4 \pi}{L} x)[ \frac{4 \pi}{L}c \times C \times -\sin( \frac{4 \pi}{L}ct) + \frac{4 \pi}{L} \times D \times \cos( \frac{4 \pi}{L} ct) ] $

$ u_t(x,0) = \frac{4 \pi}{L} c BD \sin( \frac{4 \pi}{L} x ) = 8 \pi \sin ( \frac{8 \pi}{L}x ) $

How do I solve this? Can I take $ BD = \frac{L}{c} \times 4 \cos( \frac{4 \pi}{L} x ) $

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You won't get $\sin(\frac{8\pi}{L}x)$. You are actually dealing with $\cos(\frac{4\pi}{L}c 0)=1$.

Update: I see that you may have a typo in the initial velocity condition which should have contained $\sin(\frac{8\pi}{L}x)$.

To handle that begin with the solutions as $\sum_n a_n e^{i\frac{2\pi n}{L}(x-ct)}+b_ne^{i\frac{2\pi n}{L}(x+ct)}$. With the initial conditions you can easily recognize the only coefficients left would be $a_2,b_2,a_4,b_4$. With the constraints of getting real solutions, you can see it is $\frac12 \sin(\frac{4\pi}{L}(x-ct))+\frac12\sin(\frac{4\pi}{L}(x-ct) + \frac{L}{2c}\cos(\frac{8\pi}{L}(x-ct))-\frac{L}{2c}\cos(\frac{8\pi}{L}(x+ct)).$

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