这是一份Sydney悉尼大学MATH3078/MATH3978的成功案例

Note that $c=2, f(x)=\sin x$ and $g(x)=2 \cos x$. Proceeding as before, we set
$$
f^{(2 n)}(x)=(-1)^{n} \sin x, n=0,1,2, \cdots
$$
and
$$
g^{(2 n)}(x)=2(-1)^{n} \cos x, n=0,1,2, \cdots
$$
The solution in a series form is readily obtained by substituting and given by
$$
\begin{aligned}
u(x, t)=& \sin x\left(1-\frac{(2 t)^{2}}{2 !}+\frac{(2 t)^{4}}{4 !}+\cdots\right) \
&+\cos x\left((2 t)-\frac{(2 t)^{3}}{3 !}+\frac{(2 t)^{5}}{5 !}-\cdots\right)
\end{aligned}
$$
and in a closed form by
$$
\begin{aligned}
u(x, t) &=\sin x \cos (2 t)+\cos x \sin (2 t) \
&=\sin (x+2 t)
\end{aligned}
$$

MATH3078/MATH3978 COURSE NOTES ：

To determine the function $F(x)$, we solve the second order linear ODE
$$
F^{\prime \prime}(x)+\lambda^{2} F(x)=0,
$$
to find that
$$
F(x)=A \cos (\lambda x)+B \sin (\lambda x),
$$
where $A$ and $B$ are constants. To determine the constants $A, B$, and $\lambda$, we use the homogeneous boundary conditions
$$
u(0, t)=0, u(L, t)=0,
$$
as given above gives
$$
F(0)=0, F(L)=0 .
$$
Using leads to
$$
A=0
$$
and
$$
\sin (\lambda L)=0 .
$$