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:

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


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

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)?


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 :)


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