Author Topic: Quiz 1 - P3  (Read 341 times)

Emily Deibert

  • APM346
  • Elder Member
  • *
  • Posts: 100
  • Karma: 0
    • View Profile
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}