### Author Topic: HA2-P1  (Read 3068 times)

#### Victor Ivrii

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##### HA2-P1
« on: September 28, 2015, 01:01:16 PM »

#### Emily Deibert

• Elder Member
• Posts: 100
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##### Re: HA2-P1
« Reply #1 on: October 01, 2015, 01:53:36 PM »
a)
i. We have characteristics:
\frac{dt}{2} = \frac{dx}{3}

So:

3t = 2x + C

The general solution is:

u = \phi(3t - 2x)

ii. We have characteristics:
\frac{dt}{1} = \frac{dx}{t}

So:

\frac{t^2}{2} = x + C

The general solution is:

u = \phi(\frac{t^2}{2} - x)

iii. We have characteristics:
\frac{dt}{1} = \frac{dx}{x}

So:

t = \ln(x) + C

The general solution is:

u = \phi(t - \ln(x))

iv. We have characteristics:
\frac{dt}{1} = \frac{dx}{x^2}

So:

t = \frac{-1}{x} + C

The general solution is:

u = \phi(t + \frac{1}{x})

v. We have characteristics:
\frac{dt}{1} = \frac{dx}{x^3}

So:

t = \frac{-1}{2x^2} + C

The general solution is:

u = \phi(t + \frac{1}{2x^2})

#### Victor Ivrii

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