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Learning ODE now, and using method of Undetermined Coeff

$$y'' +3y' - 7y = t^4 e^t$$

The book said that $r = 1$ is not a root of the characteristic equation. The characteristic eqtn is $r^2 + 3r - 7 = 0$ and the roots are $r = -3 \pm \sqrt{37}$

Where on earth are they getting $r = 1$ from?

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I don't understand. $r=1$ simply does not solve the equation $r^2+3r-7=0$. Hence 1 is not a root of the characteristic equation. –  Stefan Geschke Mar 1 '12 at 20:26
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@num If you don't add more information, we'll be as lost as you. I could also say $r=8$ or $r=0$ aren't roots either, but I don't know how that will help you solve your problem. –  Pedro Tamaroff Mar 1 '12 at 20:31

2 Answers 2

up vote 3 down vote accepted

$1$ comes from the $e^t$ on the right side. If it was $e^{kt}$ they would take $r=k$.

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Thanks Robert! Now I understand –  num Mar 1 '12 at 20:47

It may be related to this:

For the equation: $y'' +3y' - 7y = e^{t}$. Using the method of undetermined coefficients, you would guess that $Ae^t$ is a particular solution of the equation. But this wouldn't work if $Ae^t$ were a solution to the homogeneous equation. Then, you'd guess $Ate^t$ for a particular solution (assuming that wasn't a solution to the homogeneous equation, in which case you'd try $t^2e^t$). To check if $Ae^t$ is a solution to the homogeneous equation, you'd check if $r=1$ is a solution to the c.e..

In your case, I think, the reason for mentioning that $r=1$ is not a solution of the c.e., is because that tells you that the guess for your particular solution should contain a term $Ae^t$ (the guess contains other terms because you have $t^4e^t$ on the right hand side).

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