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Find a general solution of the PDE for $u=u(x,y)$ by using ODE techniques.

This is a simple one, but I got stuck towards the end. I wanted to use separation of variables because it was the easiest choice. Anyway, the problem is

$u_x-2u=0$

So, clearly I'm in $C^1$ for first order partial derivative and anything in $C^k$ must be continuous. Now, I rewrite the equation like this

$\frac{du}{dx}-2u=0$

Adding $2u $ to both sides, I get

$\frac{du}{dx}=2u$

Now I multiply the $dx$ and divide the $2u$

$\frac{du}{2u} = dx$

Now this is where I am stuck... I know that by integrating to both sides I get

$\frac{1}{2}ln \mid u \mid = x+C$

I feel like using an exponential log law to make it

$ln \mid u \mid^2=x+C$

and take the exponential, but the problem is that I would have

$e^{ln \mid u \mid^2}=e^{x+C}$

That's $u^2 = e^{x+C}$!!!

and I don't want that.

The answer is $u(x,y) = f(y)e^{2x}$, but I sort of am not sure where they got the $e^{2x}$ unless I'm applying the log law wrong. Also, I'm curious to know what $f(y)$ is doing here, but then again that could have something to do with $C^1$ because $f(y) $ is required to be a $C^1$ function.

usukidoll
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2 Answers2

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$\textbf{hint}$ The first step is the issue when you integrate a single variable you have freedom of a constant, however in multivariable calculus, a function that is "constant" is also $g(y)$ with respect to x, since the derivative is zero.

Chinny84
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  • so suppose I take the partial derivative of x for g(y)...would it be 0 since there is no x variable to begin with? I was also doing some practice problems for integration as well. – usukidoll Sep 09 '14 at 08:27
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    @usukidoll Partial derivative of g(y) w/rt x. Yes, it would be zero. – BCLC Sep 09 '14 at 08:28
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    Essentially yes. When you know you have a multivariable function say $u(x,y)$ then the derivative of both $x + c$ and $x +f(y)$ is 1. But restricting a knowingly multivariable fiction to a straight up constant removes the y dependence artificially and generally speaking erroneously. Unless you have good reason to think otherwise then use a the latter. :) – Chinny84 Sep 09 '14 at 08:30
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    @BCLC all right! I knew I read that section correctly... I read it slowly and followed along with the example . And then I have done some practice problems too – usukidoll Sep 09 '14 at 08:31
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The mistake is the line after "I feel like using an exponential log law to make it"

It is instead $ln |u|^{\frac{1}{2}}$. Anyway, what you should instead do is multiply both sides by 2 and raise both sides to e.

Oh also C is a function of y, not a real number. Partial integration, I think it's called.

Try it out. Or:

${\frac{1}{2}}ln |u| = x + C(y)$

$ln |u| = 2x+2C(y)$

$|u| = e^{2x+2C(y)}$

$u = +/- e^{2x+2C(y)}$

$u = f(y)e^{2x}$

where $f(y) =+/- e^{2C(y)} $

BCLC
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    ok multiplying by 2 made my life much easier because I can take the exponential on both sides to have u by itself and then by the absolute value laws I have the +/- . But suppose I keep the $\frac{1}{2}$ and take the exponential all around. Won't I have $u^{\frac{1}{2}}=e^{x+C(y)}$? Should I multiply by 2 and I have it $ u = e^{2x+2C(y)}$? – usukidoll Sep 09 '14 at 08:23
  • I believe you would then raise both sides by 2 rather than multiply by 2. – BCLC Sep 09 '14 at 08:26
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    yeah that's what I should have typed...it's getting late but yes I do mean raising both sides by 2. multiplying by 2 is way different. – usukidoll Sep 09 '14 at 08:28
  • Oh wait. Sorry. You're actually wrong. Didn't notice at first. It's supposed to be $|u|^{\frac{1}{2}} = e^{x + C(y)}$ – BCLC Sep 09 '14 at 08:30
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    $|u|^{\frac{1}{2}} = e^{x + C(y)}$ and then raise to the 2nd power? – usukidoll Sep 09 '14 at 08:36
  • @usukidoll Edit: That is correct. – BCLC Sep 09 '14 at 08:37
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    one final question. Why is $2C(y) = f(y)$? Is that something to do with the $C^k $ function? I know that the question is first order so we're in $C^1$ so I think the $f(y)$ is appropriate. – usukidoll Sep 09 '14 at 08:40
  • Damn was I that confusing? Sorry :( The equation is actually a definition: $f(y) \equiv +/- e^{2C(y)} $. The point is that u is $e^{2x}$ times some function of y given as $+/- e^{2C(y)} $. Usually textbooks eliminate the +/- from solutions of a differential equation. – BCLC Sep 09 '14 at 08:42
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    Oh no. I was just curious. You weren't confusing at all. I think it may have something to do with my book or the fact that my professor didn't do an example on it yet – usukidoll Sep 09 '14 at 08:50
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    I'm going to bed... thanks for the help and explanation :) – usukidoll Sep 09 '14 at 08:51
  • @usukidoll Even if your professor didn't do an example on it, just think of it by analogy with solving an ordinary differential equation. You get rid of +/- using constants right? – BCLC Sep 09 '14 at 08:51
  • @BCLC Oh okay. Glad to help. Try checking your ODE book though. This seems like a simple multivariate extension of separable ODE problems. – BCLC Sep 09 '14 at 08:53