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I have a uniform string of length $L$, linear mass density $\mu$, subject to tension $T$. It satisfies the wave equation:

$$\frac{\partial^2 y}{\partial t^2}=c^2\frac{\partial^2 y}{\partial x^2}$$ where $y(x,t)$ is its displacement and $c=\sqrt{\frac{T}{\mu}}$.

The system is initially at rest, one end is fixed and the other is subject to a periodic function $f(t)$, giving the below set of boundary conditions. $$y(x,0)=0; \quad \frac{\partial y(x,0)}{\partial t} = 0; \quad y(L,t)=0; \quad y(0,t) = f(t)$$

What is the solution to the wave equation in this case?

My attempt:

I have assumed $y(x,t) = X(x)T(t)$, giving

$$\frac{X''(x)}{X(x)} = \frac{T''(t)}{c^2T(t)} = \gamma $$

where $\gamma$ is some constant. This yields the following ODEs: $$T''(t) - c^2\gamma T(t) = 0 \\ X''(x) - \gamma X(x) = 0$$

If I try to solve for $X(x)$ using a characteristic equation and obtaining $\lambda^2=c^2\gamma$, then I arrive with the below possibilities depending on the value of $\gamma$:

$$X(x) = \begin{cases} Ax+B & \gamma = 0 \\ Ae^{c\sqrt{\gamma}x} + Be^{-c\sqrt{\gamma}x} & \gamma>0 \\ A\cos(\sqrt{-\gamma}x) + B\sin(\sqrt{-\gamma}x) & \gamma<0 \end{cases}$$

Here I get stuck as I don't think my boundary conditions define the coefficients in any case.

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