0
$\begingroup$

Given is an hom. ODE with constant constant coefficients:

$A_0y(x)+A_1y'(x)+A_2y''(x) + \dots + A_ny^{(n)}=0 \tag{1}$

Now it's clear to me that the solution space $y\in\mathbb L$ is a vector space.

We can solve (1) using the Ansatz: $y(x)=e^{kx}$. We get:

$\chi(k)=A_0 + A_1k + A_2k^2 + \dots + A_nk^n=0 \tag{2}$

Now with $k_i$ being a solution to $\chi$ with multiplicity $m_i$, the solution to the ODE is:

$y(x)=\sum_i y_i(x) \tag{3}$

with

$y_i(x)=\sum_{j=0}^{m_i} C_j x^j e^{k_i x}, \quad C_j\in\mathbb R \tag{4}$

Question: Since I expect $\mathbb L$ to be a vector space, (4) kind of makes sense. I mean it doesn't look wrong and I can work with it and solve such ODEs, but I can't derive it. So (1), (2), (3) is clear but (4) isn't. How exactly do we get the $x^j$ part in (4)?

$\endgroup$
1
$\begingroup$

Defining $v = (y, y^{(1)}, ..., y^{(n-1)})$, you can interpret your ODE as a linear system of differential equations,

$\frac{dv}{dt} = Av$

$v(0) = v_0$

where $A$ is an appropriate constant matrix. Now, you can show that

$v(t) = (I + At + \frac{(At)^2}{2!} + ...)v_0$

is a well defined analytic solution of this problem (in fact, the only solution). This infinite sum defines the exponential of a matrix, $e^{At}$.

Example: $y'' - 2y' + y = 0$. Defining $v = (y, y')$, we have

$v' = \begin{pmatrix} 0 & 1 \\ -1 & 2 \end{pmatrix} v$

The characteristic polynomial of this matrix (and of the associated ODE) has only one root, and for that reason the exponential of $At$ will be something of the form

$e^{At} = Qe^t\begin{pmatrix} 1 & 0 \\ t & 1 \end{pmatrix}Q^{-1}$, where Q is some constant matrix (search for Jordan Decomposition, in the context of the exponential of a matrix). And that's where the $t^j$ will come from :)

$\endgroup$

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.