Hey all

How do I find an 'explicit formula' for a solution y(x), which also satisfies initial conditions y'(0) = 1 and y(0) = 0? - where the non-linear 2nd O.D.E is

*_*_*_*_*_* MERRY CHRISTMAS*_*_*_*_*_*

Thanks in advance

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- Dec 21st 2008, 04:07 AM #1

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- Dec 21st 2008, 04:29 AM #2

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Let

And hence

You can then rearrange to get:

Which is solveable by some fairly simple integration , don't forget to sub back into y.

Explicit just means you write it in the form

This works as a general rule btw. If there is no "y" term in your 2nd order ODE, then you can transform it into a first order ODE using substitution as shown above.

- Dec 21st 2008, 05:42 AM #3

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- Dec 25th 2008, 07:43 PM #4

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- Dec 25th 2008, 08:07 PM #5

- Dec 25th 2008, 08:31 PM #6

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- Dec 25th 2008, 09:08 PM #7

- Dec 25th 2008, 10:08 PM #8
You would surely use the given boundary conditions as you go along.

It's ridiculous to carry an arbitrary constant into the second integration when it's possible to get its value. Not the least reason being that the technique required for doing the second integration will depend on what the value of that first arbitrary constant is ....

- Jan 12th 2009, 07:04 PM #9

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