Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
@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
up vote 3 down vote accepted

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

share|cite|improve this answer
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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.