# Euler-Ansatz for hom. ODEs with constant coefficients

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)?

## 1 Answer

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 :)