(A) Solve the initial-boundary value problem using separation of variables.should be provided.The details of the derivations \begin{aligned} \frac{\partial^{2} y}{\partial t^{2}} &=2 \frac{\partial^{2} y}{\partial x^{2}} \text { for } 0<x<4, t>0 \\ y(0, t) &=y(4, t)=0 \text { for } t \geq 0 \\ y(x, 0) &=0, \frac{\partial y}{\partial t}(x, 0)=x(4-x) \text { for } 0 \leq x \leq 4 \end{aligned} \begin{aligned} \frac{\partial^{2} y}{\partial t^{2}} &=16 \frac{\partial^{2} y}{\partial x^{2}} \text { for } 0<x<\pi / 3, t>0 \\ y(0, t) &=y(\pi / 3, t)=0 \text { for } t \geq 0 \\ y(x, 0) &=\sin (3 x), \frac{\partial y}{\partial t}(x, 0)=\pi / 3-x \text { for } 0 \leq x \leq \pi / 3 \end{aligned} \begin{aligned} \frac{\partial^{2} y}{\partial t^{2}} &=36 \frac{\partial^{2} y}{\partial x^{2}} \text { for } 0<x<8, t>0 \\ y(0, t) &=y(8, t)=0 \text { for } t \geq 0 \\ y(x, 0) &=2 \sin (\pi x), \frac{\partial y}{\partial t}(x, 0)=0 \text { for } 0 \leq x \leq 8 \end{aligned} \begin{aligned} \frac{\partial^{2} y}{\partial t^{2}} &=25 \frac{\partial^{2} y}{\partial x^{2}} \text { for } 0<x<\pi, t>0 \\ y(0, t) &=y(\pi, t)=0 \text { for } t \geq 0 \\ y(x, 0) &=\sin (x / 2), \frac{\partial y}{\partial t}(x, 0)=x \text { for } 0 \leq x \leq \pi \end{aligned} \begin{aligned} \frac{\partial^{2} y}{\partial t^{2}} &=2 \frac{\partial^{2} y}{\partial x^{2}} \text { for } 0<x<\pi, t>0 \\ y(0, t) &=y(\pi, t)=0 \text { for } t \geq 0 \\ y(x, 0) &=\cos (x), \frac{\partial y}{\partial t}(x, 0)=x^{2} \text { for } 0 \leq x \leq \pi \end{aligned}

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