Using exponential substitution:

-\infty,0]\cup [1,\infty)" alt=" \ \ s-\infty,0]\cup [1,\infty)" />

Can this be simplified more or is there a more tactful way to represent this?

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- Dec 29th 2010, 07:53 PM #1

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## u_{xx}-u_{yy}+u_y=0

Using exponential substitution:

-\infty,0]\cup [1,\infty)" alt=" \ \ s-\infty,0]\cup [1,\infty)" />

Can this be simplified more or is there a more tactful way to represent this?

- Dec 29th 2010, 09:11 PM #2

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- Dec 29th 2010, 10:34 PM #3

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- Dec 29th 2010, 10:39 PM #4

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- Dec 29th 2010, 10:41 PM #5

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- Dec 30th 2010, 04:39 AM #6

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Dear dwsmith,

I have taken so that, I get required value for s;

If I didnt know the answer I would not present it as I had done in post #3. I would have given the answer as,

But since you had given thes answers in the book and told that you are not sure how they have been obtained I had shown that both anwers are equivelent. Does this clarify your doubts?

- Dec 30th 2010, 12:33 PM #7

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Now I am solving the same equation using separation of variables. However, I have hit a snag.

The first equation is suppose to yield , and the second equation needs to be of the form in order to obtain the same solutions from the exponential method.

I haven't come across this in the book but can I just set both equations equal to

- Dec 30th 2010, 04:15 PM #8

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Dear dwsmith,

Yes you can. What is the reasoning behind setting ?? It is that the two parts are independent of y and x respectively. So that if they are equal to each other there must be a common constant to which both are equal. In this case you had taken it to be . But you can take the constant as you like it to be. For example instead of you can take if you dont like taking a the negative sign. Then there exists such that . Substitute for and you will get the same answer.

Hope you got the idea. I am a non native speaker of english. So if you find anything that confuses you please do not hesitate to ask me.