Higher power constant coefficiants ODEs

[ognik]

Hi, I have the ODE y'''' - 3y' + 2y = 0

The characteristic equation is then $r^4 - 3r + 2 = 0$

So my 1st question, is there some easier way of finding the roots than long division?

I looked at the first and last terms to guess the roots (if real) might come from (r-1) , (r+1), (r+2), (r-2)

(r-1) divides into the CE with remainder $ r^3+r^2+r-2 $ but none of my other guessed terms divide into this exactly, so I am thinking the other roots must be complex ... so how would I go about finding the rest of the roots please?

[I did try multiplying out (r-1)(r+a)(r+b)(r+c) and comparing coefficients, but the eqtns I got were not independent so couldn't solve for a,b,c]

 

[Ackbach]

There's always synthetic division. Also, I'm not sure you're using the Rational Roots Theorem properly. [EDIT] You were. See below for correction. Your rational root candidates would be factors of $1$ divided by factors of $2$. So you should test $\pm 1, \pm \dfrac12$. If you look at the coefficients, they sum to zero, indicating that $r=1$ is a root, or $r-1$ is a factor. Using synthetic division yields the factored form
$$(r-1)\left(r^3+r^2+r-2\right)=0.$$
I'm not sure you'll find any more rational roots. There's one more positive root (by Descarte's Rule of Signs). Replacing $r$ with $-r$ in the cubic yields $-r^3+r^2-r-2=0$, with two sign changes. So you either have two negative real roots, or no negative real roots (the latter would imply you have two complex conjugate roots). At this point, I'd probably go with Mathematica to get an exact answer, or you can simply go numerical if the exact answer is not critical.

 

[MarkFL]

...Also, I'm not sure you're using the Rational Roots Theorem properly...

He did it correctly...you want the integer factors of the trailing term in the numerator and the integer factors of the leading term in the denominator. :)