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2nd order Linear DE

 
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Jan17-13, 12:28 PM   #1
 

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?
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Jan17-13, 02:33 PM   #2
 
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Hi Apteronotus!
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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