Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.


Topics - Emily Deibert

Pages: 1 [2]
16
Textbook errors / Broken link in web bonus problem week 4 #6
« on: October 08, 2015, 03:37:58 PM »

17
Textbook errors / Typo in HA3 P3?
« on: October 03, 2015, 05:54:34 PM »
I asked this in the board for HA3 P3 as well but I am cross-posting here.

Quote
Just wondering, is there a typo in the original problem? The equation given is: \begin{equation}
Au_{tt} + 2Bu_{tx} + Cu_{tt} \end{equation}
But I think the last term should be with respect to x: \begin{equation}
Au_{tt} + 2Bu_{tx} + Cu_{xx}
\end{equation}
Indeed, corrected

18
Quiz 1 / Quiz 1 - P3
« on: October 02, 2015, 12:35:31 AM »
This problem was: \begin{equation} \begin{cases}
u_x + 3u_y = xy \\
u|_{x=0} = 0
\end{cases}
\end{equation}
And my solution is: \begin{equation}
\frac{dx}{1} = \frac{dy}{3} = \frac{du}{xy}
\end{equation}

\begin{equation}
3dx = dy
\end{equation}

\begin{equation}
3x - y = C
\end{equation}

\begin{equation}
du=xydx
\end{equation}

\begin{equation}
du = x(3x-C)dx
\end{equation}

\begin{equation}
du = (3x^2-Cx)dx
\end{equation}

\begin{equation}
u = x^3 - \frac{C}{2}x^2 + \phi(3x-y)
\end{equation}

\begin{equation}
u = x^3 - \frac{3x-y}{2}x^2 + \phi(3x-y)
\end{equation}

With initial condition, we have:
\begin{equation}
u|_{x=0}=\phi(-y)=0
\end{equation}

So: \begin{equation} \phi = 0 \end{equation}

So the final solution will be: \begin{equation}
u = x^3 - \frac{3x-y}{2}x^2
\end{equation}

19
Textbook errors / Error in HA 2 Problem 1?
« on: October 01, 2015, 01:22:09 PM »
Hi Professor,

HA2 Problem 1 a) has as the fifth problem, \begin{equation}
u_x + x^3u_x = 0
\end{equation}

Did you perhaps mean \begin{equation}
u_t + x^3u_x = 0
\end{equation}

to make an equation involving partial derivatives with respect to t AND x, like the other problems?


Pages: 1 [2]