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# How do I solve this 1st order linear PDE Watch

1. Hi, I'm new to this forum, just to be able to ask you this question.

I have a PDE that I want to solve and has been looking for a solution through the Method of Characteristics, but fail to complete one of the steps.

Here is my equation:
d(u(x,t))/dt = (ax^2-(a+b)x+b)d(u(x,t))/dx

don't know how to write it in math characters but maybe you could,
here is a picture of it though:

Thanks! // Marcus from Sweden
2. Oh I found out you could write in LaTeX, so here it is again:

[latex]
\frac{\partial}{\partial t}u(x,t) = (ax^2-(a+b)x+b) \frac{\partial}{\partial x}u(x,t)
[\latex]

And when I started with MOC, and created my new variables s,r

I use the chainrule and get:
[latex] \frac{du}{dr} = u_t \frac{dt}{dr} + u_x \frac{dx}{dr}
[\latex]
which is my initial PDE, if

[latex] \frac{dt}{dr} = 1 [\latex] and [latex] \frac{dx}{dr} = ax^2-(a+b)x+b
[\latex]
but that last equation is quite a challenge, or is it?
..because x in this case depend on r and s, right?

3. Why didn't the Tex code output what it should have??
4. The slashes for the Latex tags are the opposite way around (plus there are some other bugs). Here's a fixed version.

(Original post by vidomir)
Oh I found out you could write in LaTeX, so here it is again:

And when I started with MOC, and created my new variables s,r

I use the chainrule and get:

which is my initial PDE, if

and

but that last equation is quite a challenge, or is it?
..because x in this case depend on r and s, right?

5. (Original post by vidomir)
Hi, I'm new to this forum, just to be able to ask you this question.

I have a PDE that I want to solve and has been looking for a solution through the Method of Characteristics, but fail to complete one of the steps.

Here is my equation:
d(u(x,t))/dt = (ax^2-(a+b)x+b)d(u(x,t))/dx

don't know how to write it in math characters but maybe you could,
here is a picture of it though:

Thanks! // Marcus from Sweden
From equation

that is

Integrating

and simplify
use

for |y|<1
6. Thanks ztibor, I wasn't able to see your answer until today and will have no time to go through it right now but I will definitely take a look at it soon. Seems to be the way to go with that.

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Updated: April 13, 2013
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