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2nd order Linear DE |
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| Jan17-13, 12:28 PM | #1 |
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2nd order Linear DE
Hi,
When solving a 2nd order Linear DE with constant coefficients ([itex]ay''+by'+cy=0[/itex]) we are told to look for solutions of the form [itex]y=e^{rt}[/itex] and then the solution (if we have 2 distinct roots of the characteristic) is given by [itex]y(t)=c_1 e^{r_1 t}+c_2 e^{r_2 t}[/itex] This is clearly a solution, but how do we know there are no other solutions? That is, how do we know this is the general solution? |
| Jan17-13, 02:33 PM | #2 |
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Hi Apteronotus!
![]() if y' - ry = 0, put y = zert, then (z' + rz)ert = rzert so ert = 0 (which is impossible), or z' + rz = rz, ie z' = 0, ie z is constant ![]() and now try (y' - ry)(y' - sy) = 0, using the same trick twice |
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