---
tag: 0HJ0HDA6W7
kind: statement
role: theorem
name: Solutions for distinct real roots
msc: [34A30]
uses: [0QDXMG46QW, 0H4KNNCHPY, 0A6PGTTP2A]
sources:
  - url: https://en.wikipedia.org/wiki/Characteristic_equation_(calculus)
    locator: Distinct real roots
---

## Statement

Let $a_0, \dots, a_{n-1}$ be real numbers, and suppose that the [[characteristic polynomial|0QDXMG46QW]] $p(r) = r^n + a_{n-1} r^{n-1} + \dots + a_0$ 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|0H4KNNCHPY]] 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|0MHE4B1QCQ]] 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.