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- January 4th 2011, 05:15 PM #1

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- January 4th 2011, 07:45 PM #2

- January 4th 2011, 08:09 PM #3

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- January 5th 2011, 07:19 AM #4

- January 5th 2011, 07:35 AM #5

- January 5th 2011, 12:58 PM #6

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- January 5th 2011, 01:00 PM #7

- January 5th 2011, 05:47 PM #8
Here's my two cents on how to do this problem. I will be skipping some steps; you should be able to fill them in without a problem.

I would separate variables right away. Let (for short, ).

So we end up with

Since each side of the equation is defined in a different variable when compared to one another, we can conclude that LHS = RHS only when they're equal to the same constant, call it .

Thus, we can rewrite this as two different ODEs:

Consider the first equation. To make life easier, let . Then we see that the equation becomes

(*)

As TES mentioned in his post, you're going to get different solutions depending on the value of . I will do the case and leave the other two cases for you to work out.

So the characteristic equation for (*) is . Assuming , we see that . So, we see that .

But .

Now, we do something similar for the second equation. The nice thing is that this equation is**separable**:

So we see that (aside question: is it ok to remove absolute values at this point?)

Thus, your solution should be .

Reiterating what TES said, you will get other solutions for testing the other cases.

Hopefully this makes some sense.

- January 5th 2011, 05:50 PM #9

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