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Do the roots of the characteristic polynomial must be integers?

For example for the ode $u^{(5)}+5u^{(4)}-2u^{(3)}-10u^{(2)}+u'+5u=0$

The characteristic polynomial is $\lambda^{(5)}+5\lambda^{(4)}-2\lambda^{(3)}-10\lambda^{(2)}+\lambda+5=0$

  1. Why for $u'$ we take $\lambda$ and why for $5u$ we take $5?

  2. To find the roots, I can first "guess" one root and divided the characteristic polynomial to find the rest, how can I be sure that $\lambda\in \mathbb{Z}$?

3 Answers 3

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These aren't derivate but are powers: $$\lambda^5+5\lambda^4-2\lambda^3-10\lambda^2+\lambda+5=0$$ $$(\lambda^5-2\lambda^3+\lambda)+5(\lambda^4-2\lambda^2+1)=0$$ $$\lambda(\lambda^4-2\lambda^2+1)+5(\lambda^4-2\lambda^2+1)=0$$ and continue ...

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First of all, there's a typo when you wrote $\lambda^{(5)}$ because what it should say is $\lambda^5$. It's simply a power of $\lambda$, there are no derivatives involved. Same for $\lambda^{(4)},\lambda^{(3)},\lambda^{(2)}$.

No, the roots of the characteristic polynomial do not need to be integers. For example, the differential equation $$2u' + u = 0$$

has the characteristic polynomial $2\lambda +1=0$.

Your other questions:

  1. For $u'$, we take $\lambda$, because the standard method of solving linear ODEs is to replace $u^{(k)}$ with $\lambda^k$ for each value of $k$. Applying this rule means we replace $u'=u^{(1)}$ with $\lambda^1=\lambda$, and we replace $u=u^{(0)}$ with $\lambda^0=1$.
  2. There is no way to solve a general polynomial equation $p(\lambda)=0$. You can use all the tricks you learnt in high school, (i.e., it's easy if $p$ has degree $2$, and you can write down the rational candidates using Eisenstein's criterion https://en.wikipedia.org/wiki/Eisenstein's_criterion) but there will always be a chance that the solution can only be approximated numerically - they can be integers, sure, but they can also be irrational numbers.
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Since characteristic polynomials are, in fact (you guessed it) polynomials, their roots can be every number.

Since you can get a polynomial by wishing some roots and use linear factors to build it from there, you can get a linear differential equation with every root you wish.

The coefficients might not be integer-values than. Keep in mind, that no closed forms for roots exist foo most higher order polynomials, therefore making it impossible to get a analytic solution, if the roots are "anywhere". That would not make a very good textbook, would it?