Timeline for Does it mean that there no analytical solution if Mathematica cannot find analytical solution?
Current License: CC BY-SA 3.0
5 events
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| Mar 25, 2016 at 18:48 | comment | added | MathX |
oh yes you are right. And in that case it is equal to a from the OP.
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| Mar 25, 2016 at 18:41 | history | edited | Michael Seifert | CC BY-SA 3.0 |
added 49 characters in body
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| Mar 25, 2016 at 18:40 | comment | added | Michael Seifert |
$c_3$ can be determined from the initial conditions: since it's a constant, then $c_3 = x(0) + y(0)$. I've edited to clarify this. In general, you'll have two constants of integration for this system; in my solution, they'll be the $c_3$ I coded in and the C[1] generated by Mathematica in the solution.
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| Mar 25, 2016 at 18:39 | comment | added | MathX | Wait, but now don't we have an unknown c3 in our solution? Meaning, x[t] is now expressed in terms of a y[t] that we still don't know what it is. | |
| Mar 25, 2016 at 17:34 | history | answered | Michael Seifert | CC BY-SA 3.0 |