Author Topic: HA2-P1  (Read 3068 times)

Victor Ivrii

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Emily Deibert

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Re: HA2-P1
« Reply #1 on: October 01, 2015, 01:53:36 PM »
a)
i. We have characteristics: \begin{equation}
\frac{dt}{2} = \frac{dx}{3}
\end{equation}
So:
\begin{equation}
3t = 2x + C
\end{equation}
The general solution is:
\begin{equation}
u = \phi(3t - 2x)
 \end{equation}


ii. We have characteristics: \begin{equation}
\frac{dt}{1} = \frac{dx}{t}
\end{equation}
So:
\begin{equation}
\frac{t^2}{2} = x + C
\end{equation}
The general solution is:
\begin{equation}
u = \phi(\frac{t^2}{2} - x)
 \end{equation}


iii. We have characteristics: \begin{equation}
\frac{dt}{1} = \frac{dx}{x}
\end{equation}
So:
\begin{equation}
t = \ln(x) + C
\end{equation}
The general solution is:
\begin{equation}
u = \phi(t - \ln(x))
 \end{equation}


iv. We have characteristics: \begin{equation}
\frac{dt}{1} = \frac{dx}{x^2}
\end{equation}
So:
\begin{equation}
t = \frac{-1}{x} + C
\end{equation}
The general solution is:
\begin{equation}
u = \phi(t + \frac{1}{x})
 \end{equation}


v. We have characteristics: \begin{equation}
\frac{dt}{1} = \frac{dx}{x^3}
\end{equation}
So:
\begin{equation}
t = \frac{-1}{2x^2} + C
\end{equation}
The general solution is:
\begin{equation}
u = \phi(t + \frac{1}{2x^2})
 \end{equation}

Victor Ivrii

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Re: HA2-P1
« Reply #2 on: October 03, 2015, 05:46:10 AM »
There are more complicated parts but you need to look how characteristics go. In two last problems they escape to infinity by $x$ for a finite time $t$ but still each of them strikes $t=0$ once