Toronto Math Forum

Hi, I just want to confirm with someone about the solution for Term test 2 variant E problem2. I think for$\lambda_n$ the n should start from 0, so n should be n = 0, 1, 2 and so on. I don't know if the solution makes a typo or am I missing some insights here.

For problem 1 in Practice Test Variant E, I'm wondering if this is a typo because the interval given is from $-\pi$ to $\pi$? (I circled it in red)

\begin{equation}
    \begin{cases}
      u_{tt}- c^2u_{xx} = 0 &\ x>0\\
      u\rvert_{t = 0} = sech(x) &\ x>0\\
      u_{t}\rvert_{t = 0} = 0 &\ x>0\\
      u_{x}\rvert_{x = 0} = 0 &\ t>0\\
    \end{cases}       
\end{equation}

For this problem, my approach was to extend $sech(x)$ as an even function for $x \in R$. Since $sech(x)$ is already an even function so we can write the set of equations as follows.

\begin{equation}
    \begin{cases}
      u_{tt}- c^2u_{xx} = 0 &\ x>0\\
      u\rvert_{t = 0} = sech(x)\\
      u_{t}\rvert_{t = 0} = 0 &\ x>0\\
    \end{cases}       
\end{equation}

And by using D'Alembert Formula $u(x,t) =\frac{1}{2}[g(x+ct)+g(x-ct)] $ we can simplify it as $\frac{1}{2}[sech(x+ct)+sech(x-ct)]$ $,x>0$

\begin{align}
&\left\{\begin{aligned}
&u_{xy}=0,\\[2pt]
&u_{xz}=0;
\end{aligned}\right.\\[2pt]
\end{align}

For this problem I'm not sure if my approach was correct.
\begin{align}
u_{x}=f(x,z)  \textrm{   from the first equation}
\end{align}
\begin{align}
&u_{x}=g(x,y)\textrm{   from the second equation}
\end{align}
\begin{align}
\textrm{Therefore we must have   }\textrm{   }u_{x} = h(x)
\end{align}
\begin{align}
&u=H(x) + m(y,z)\textrm{ } \textrm{where}\textrm{ } H'(x)=h(x)
\end{align}