0
$\begingroup$

If I have solution basis , how can I find out if there is a real constant coefficient homogeneous linear ODE to that solution or not?

For example, I have the solution basis $e^x, xe^x, x^2e^x,\cos(x), \sin(x)$.

I know this site is not for homework, but I can't figure it out from the book I have.

$\endgroup$
0
$\begingroup$

For each one, you have to find a constant-coefficient differential operator that eliminates it, and then you can stack them together (i.e. compose them), because if the coefficients are constant, the operators commute. For example, $\cos{x}$ and $\sin{x}$ are annihilated by $(d/dx)^2+1$. The other three can be eliminated by the same operator, $(d/dx-1)^3$: $$ \left(\frac{d}{dx}-1\right) x^2e^x = 2xe^x + x^2e^x-x^2e^x = 2xe^x, \\ \left(\frac{d}{dx}-1\right) xe^x = e^x + xe^x-xe^x = e^x \\ \left(\frac{d}{dx}-1\right) e^x = e^x-e^x = 0 $$ Hence all five are eliminated by $ \left(\frac{d}{dx}-1\right)^3 \left( \frac{d^2}{dx^2}+1 \right) $, and you can find the actual form of the differential equation by applying this to a function $y$ and expanding out.

$\endgroup$
0
$\begingroup$

Each homogenous ODE with constant coefficients can be solved by finding the roots of the characteristic polynomial, e.g. $$y'' - 3y' + 2y = 0$$ has the characteristic polynomial $r^2-3r+2=(r-1)(r-2)$ and is solved by $y = c_1e^t+c_2e^{2t}$. If you had the functions $e^t$ and $e^{2t}$ as the basis functions for the solutions, then you would know that the characteristic polynomial had roots $1$ and $2$. Thus we can build an ODE which has the given solutions if we have the basis functions (as long as they are consistent).

So if we have $$e^{st}, \\ te^{st},\\ \cdots, \\ t^{k-1}e^{st}$$ as basis functions, we would know that our characteristic polynomial has a root of $s$ with multiplicity $k$, i.e., $p(r) =(s-r)^kq(r)$ where $p(r)$ is our characteristic polynomial and $q(r)$ is another polynomial. If we have $$e^{\alpha t}\cos(\beta t), e^{\alpha t}\sin(\beta t), \\ e^{\alpha t}t\cos(\beta t), e^{\alpha t}t\sin(\beta t), \\ \cdots, \\ e^{\alpha t} t^{k-1}\cos(\beta t), e^{\alpha t}t^{k-1}\sin(\beta t)$$ as basis functions, the characteristic polynomial would have a root $\alpha + \beta i$ and $\alpha - \beta i$ of multiplicity $k$.

To take your example, $e^x,xe^x,x^2e^x$ correspond to a root of $1$ with multiplicity $3$ and $\cos(x), \sin(x)$ correspond to roots $i$ and $-i$. A candidate characteristic polynomial would be $$(r-1)^3(r-i)(r+i) = r^5-3r^4+4r^3-4r^2+3r -1.$$ And so our ODE would be $$y^{(5)}-3y^{(4)}+4y^{(3)}-4y''+3y' -y=0.$$

Note that this would not work for any choice of basis functions. For instance, there isn't a constant coefficient homogenous ODE with $te^t$ as its only basis function (we need $e^t$ as well). We also can't have just $\cos(t)$ as a basis function (we need $\sin(t)$ as well).

$\endgroup$
  • $\begingroup$ thanks a lot, but how do you find the roots? and how did you figure out that it has a multiplisity of 3 for the first root? $\endgroup$ – i.ul Feb 26 '17 at 22:22
  • $\begingroup$ @i.ul I basically ran the method of solving the ODE in reverse. If we have to solve $y''-3y'+2y=0$, we make the guess that the solution is like $y=e^{rt}$. Substituting that into the ODE gives $r^2e^{rt} -3re^{rt} +2e^{rt} = 0$ and dividing out $e^{rt}$ gives $r^2 - 3r +2=0$. Thus if $r$ is a root of the polynomial, $e^{rt}$ is a solution. When constructing the ODE from the functions, we know since $e^{rt}$ is a solution, $r$ must be a root of the polynomial. So $e^{2t}$ would be associated with a root of $2$. Figuring out the multiplicity is also just working in reverse. $\endgroup$ – Trevor Norton Feb 26 '17 at 23:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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