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 $\lambda^{(5)}+5\lambda^{(4)}-2\lambda^{(3)}-10\lambda^{(2)}+\lambda+5=0$
Why for $u'$ we take $\lambda$ and why for $5u$ we take $5?
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
$\begingroup$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 ...
$\endgroup$ $\begingroup$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:
- 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 $\lambda^0=1$.
- 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 '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.
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?
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