\[\begin{cases}
& \dfrac{\partial^2u}{\partial t^2}-a^2\dfrac{\partial^2u}{\partial x^2} = A_0\sin\omega t & 0\lt x\lt l,t\gt 0\\
& u\big\vert_{x=0}=u\big\vert_{x=l}=0 & t\ge 0\\
& u\big\vert_{t=0}=\dfrac{\partial u}{\partial t}\big\vert_{t=0}=0 & 0\le x\le l
\end{cases}
\]
设 \(u=v+w\)，其中 \(v\) 满足
\[\begin{cases}
& \dfrac{\partial^2v}{\partial t^2}-a^2\dfrac{\partial^2v}{\partial x^2} = A_0\sin\omega t & 0\lt x\lt l,t\gt 0\\
& v\big\vert_{x=0}=v\big\vert_{x=l}=0 & t\ge 0\\
\end{cases}
\]
\(w\) 满足
\[\begin{cases}
& \dfrac{\partial^2w}{\partial t^2}-a^2\dfrac{\partial^2w}{\partial x^2} = 0 & 0\lt x\lt l,t\gt 0\\
& w\big\vert_{x=0}=w\big\vert_{x=l}=0 & t\ge 0\\
& w\big\vert_{t=0}=-v\big\vert_{t=0}, \dfrac{\partial w}{\partial t}\big\vert_{t=0}=-\dfrac{\partial v}{\partial t}\big\vert_{t=0} & 0\le x\le l
\end{cases}
\]
首先求解 \(v\)，观察到条件左侧均为二阶导，可设 \(v\left(x,t\right)=f\left(x\right)\sin\omega t\)，则 \(f\left(x\right)\) 满足
\[\omega^2 f\left(x\right) + a^2 f''\left(x\right) = A_0
\]
解得 \(f\left(x\right) = -\dfrac{A_0}{\omega^2} + A\sin\dfrac{\omega}{a}x + B\cos \dfrac{\omega}{a}x\)
带入边界条件 \(f\left(0\right)=f\left(l\right)=0\) 得 \(A = \dfrac{A_0}{\omega^2}\tan\dfrac{\omega l}{2a}, B = \dfrac{A_0}{\omega^2}\)（\(\tan\dfrac{\omega l}{2a}\) 不存在时取 \(A=0\) 即可，接下来讨论 \(\dfrac{\omega l}{a\pi}\not\in\mathbb{Z}\) 的情况）
则 \(f\left(x\right) = \dfrac{A_0}{\omega^2}\left(\tan\dfrac{\omega l}{2a}\sin\dfrac{\omega}{a}x + \cos\dfrac{\omega}{a}x - 1\right) = \dfrac{A_0}{\omega^2}\left(\dfrac{\cos \dfrac{\omega}{a}\left(x-\dfrac{l}{2}\right)}{\cos\dfrac{\omega l}{2a}}-1\right)\)
再求解 \(w\)。分离变量法，设 \(w\left(x,t\right)=X\left(x\right)T\left(t\right)\)，则有
\[\dfrac{T''\left(t\right)}{T\left(t\right)} = a^2\dfrac{X''\left(x\right)}{X\left(x\right)} = -\lambda
\]
其中 \(\lambda\in\mathbb{C}\) 为与 \(x,t\) 均无关的本征值。
\(\lambda = 0\) 时，\(X''\left(x\right) = 0\)，则 \(X\left(x\right)\) 为一次函数，又由边界条件得 \(X\left(x\right)\equiv 0\)。
\(\lambda\neq 0\) 时，解得
\[X\left(x\right) = A\sin\sqrt{\lambda}x + B\cos\sqrt{\lambda}x
\]
\[T\left(t\right) = C\sin\sqrt{\lambda}at + D\cos\sqrt{\lambda}at
\]
带入 \(X\left(0\right)=X\left(l\right)=0\) 得 \(B=0, \lambda_n = \left(\dfrac{n\pi}{l}\right)^2, n\in\mathbb{N}^{\star}\)
将所有本征函数累加得
\[w\left(x,t\right) = \sum\limits_{n=1}^{\infty}\left(C_n\sin\dfrac{n\pi}{l}at + D_n\cos\dfrac{n\pi}{l}at\right)\sin\dfrac{n\pi}{l}x
\]
带入 \(v\big\vert_{t=0}\) 的初始条件，由 \(w\big\vert_{t=0}=-v\big\vert_{t=0}=0\) 可得 \(\forall n\in\mathbb{N}^{\star},D_n=0\)
由 \(\dfrac{\partial w}{\partial t}\big\vert_{t=0}=-\dfrac{\partial v}{\partial t}\big\vert_{t=0}=-\omega f\left(x\right)\) 可得 \(\forall n\in\mathbb{N}^{\star}\) 有
\[\begin{aligned}
C_n &= \dfrac{2}{l}\cdot\dfrac{l}{n\pi a}\cdot\left(-\omega\right)\cdot\dfrac{A_0}{\omega^2}\int_0^l \left(\dfrac{\cos\dfrac{\omega}{a}\left(x-\dfrac{l}{2}\right)}{\cos\dfrac{\omega l}{2a}}-1\right)\sin\dfrac{n\pi}{l}x\mathrm{d}x\\
&= -\dfrac{A_0}{n\pi a\omega l}\left(1-\left(-1\right)^n\right)\sin\dfrac{n\pi}{2}\int_{-l/2}^{l/2}\left(\dfrac{\cos\dfrac{\omega}{a}x}{\cos\dfrac{\omega l}{2a}}-1\right)\cos\dfrac{n\pi}{l}x\mathrm{d}x\\
&= -\dfrac{A_0}{n\pi a\omega l}\left(1-\left(-1\right)^n\right)\sin\dfrac{n\pi}{2}\left(\dfrac{1}{\cos\dfrac{\omega l}{2a}}\cdot\dfrac{\dfrac{2\omega}{a}\sin\dfrac{\omega l}{2a}\cos\dfrac{n\pi}{2} - \dfrac{2n\pi}{l}\cos\dfrac{\omega l}{2a}\sin\dfrac{n\pi}{2}}{\left(\dfrac{\omega}{a}\right)^2-\left(\dfrac{n\pi}{l}\right)^2}-\dfrac{2l}{n\pi}\sin\dfrac{n\pi}{2}\right)\\
&= \dfrac{A_0}{n\pi a\omega l}\left(1-\left(-1\right)^n\right)\left(\dfrac{\dfrac{2n\pi}{l}\cos\dfrac{\omega l}{2a}}{\left(\dfrac{\omega}{a}\right)^2-\left(\dfrac{n\pi}{l}\right)^2}+\dfrac{2l}{n\pi}\right)\\
&= \dfrac{1-\left(-1\right)^n}{n^2}\cdot\dfrac{2\omega A_0 l^3}{\pi^2 a}\cdot\dfrac{1}{\omega^2 l^2-n^2\pi^2a^2}
\end{aligned}
\]