Timeline for Is this 1974 claim still valid?
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| Feb 5, 2011 at 23:10 | vote | accept | Unknown | ||
| Feb 5, 2011 at 12:40 | comment | added | Willie Wong | @Elohemahab: not just $y+z$. By the definition of $(a,b)$, Denis's construction admits any linear combination (a 2-parameter family) of the functions $y$ and $z$. | |
| Feb 5, 2011 at 9:23 | history | edited | Unknown | CC BY-SA 2.5 |
added a link to the main book containg quote
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| Feb 4, 2011 at 21:06 | answer | added | agt | timeline score: 12 | |
| Feb 4, 2011 at 20:41 | answer | added | Michael Renardy | timeline score: 3 | |
| Feb 4, 2011 at 20:41 | answer | added | Igor Rivin | timeline score: 29 | |
| Feb 4, 2011 at 18:11 | comment | added | Unknown | @Anthony, the sought-after type is a 2nd order linear ODE which is not one with constant coefficients, or one reducible to such by changes of the independent variable. | |
| Feb 4, 2011 at 17:49 | comment | added | Unknown | @Denis, thanks for the edit. Is your $f$ equal to $y+z$? | |
| Feb 4, 2011 at 17:36 | comment | added | Anthony Quas | maybe it just depends on what's meant by a type of differential equation? | |
| Feb 4, 2011 at 17:28 | comment | added | Denis Serre | I don't understand the claim. Choose two elementary functions $y(t)$ and $z(t)$, functionally independent. Then the system formed by $a(t)y'(t)+b(t)y(t)+y''(t)=0$ and $a(t)z'(t)+b(t)z(t)+z''(t)=0$ admits a unique solution $(a,b)$. Then the linear 2nd-order ODE $f''+af'+bf=0$ is solvable in terms of elementary functions. What is wrong ? | |
| Feb 4, 2011 at 17:24 | history | edited | Denis Serre | CC BY-SA 2.5 |
edited body
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| Feb 4, 2011 at 17:12 | history | asked | Unknown | CC BY-SA 2.5 |