2nd order Linear DE


by Apteronotus
Tags: linear, order
Apteronotus
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#1
Jan17-13, 12:28 PM
P: 196
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?
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tiny-tim
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#2
Jan17-13, 02:33 PM
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P: 26,167
Hi Apteronotus!
Quote Quote by Apteronotus View Post
how do we know there are no other solutions?
It's easy to prove for the first-order case

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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