Solutions for distinct real roots

Statement

Let $a_0, \dots, a_{n-1}$ be real numbers, and suppose that the characteristic polynomial $p(r) = r^n + a_{n-1} r^{n-1} + \dots + a_0$ has $n$ distinct real roots $r_1, \dots, r_n$. Then $\left(e^{r_1 x}, \dots, e^{r_n x}\right)$ is a fundamental system of the equation $y^{(n)} + a_{n-1} y^{(n-1)} + \dots + a_0 y = 0$ on $\mathbb{R}$. So the solutions are exactly the functions $y = c_1 e^{r_1 x} + \dots + c_n e^{r_n x}$ with real constants $c_1, \dots, c_n$.

Remarks

Each $e^{r_k x}$ is a solution because $p(r_k) = 0$. The Wronskian at $0$ is the Vandermonde determinant $\prod_{j < k} (r_k - r_j)$, which is not zero. So the Wronskian test gives independence. Example: $y'' - 3 y' + 2 y = 0$ has the solutions $c_1 e^{x} + c_2 e^{2x}$.

Explanations

L1 human

Each root of the characteristic polynomial gives one exponential solution. If the $n$ roots are different, these $n$ exponentials are independent, and their combinations are all the solutions.