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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- December 21st 2008, 04:07 AMtsal152nd O.D.E.
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 - December 21st 2008, 04:29 AMMush
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. - December 21st 2008, 05:42 AMtsal15
- December 25th 2008, 07:43 PMtsal15
- December 25th 2008, 08:07 PMJester
- December 25th 2008, 08:31 PMtsal15
- December 25th 2008, 09:08 PMJester
- December 25th 2008, 10:08 PMmr fantastic
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 .... - January 12th 2009, 07:04 PMtsal15
Finding the general solution of the 2ODE, was the process we have been discussing so far?