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  "body": "\n## Statement\n\nLet $a_0, \\dots, a_{n-1}$ 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 $r_1, \\dots, r_n$.\nThen $\\left(e^{r_1 x}, \\dots, e^{r_n x}\\right)$ is a [[fundamental system|0H4KNNCHPY]] of the equation $y^{(n)} + a_{n-1} y^{(n-1)} + \\dots + a_0 y = 0$ on $\\mathbb{R}$.\nSo the solutions are exactly the functions $y = c_1 e^{r_1 x} + \\dots + c_n e^{r_n x}$ with real constants $c_1, \\dots, c_n$.\n\n## Remarks\n\nEach $e^{r_k x}$ is a solution because $p(r_k) = 0$. The Wronskian at $0$ is the Vandermonde determinant $\\prod_{j \u003c k} (r_k - r_j)$, which is not zero. So the [[Wronskian test|0MHE4B1QCQ]] gives independence.\nExample: $y'' - 3 y' + 2 y = 0$ has the solutions $c_1 e^{x} + c_2 e^{2x}$.\n\n## Explanations\n\n### L1 human\n\nEach root of the characteristic polynomial gives one exponential solution.\nIf the $n$ roots are different, these $n$ exponentials are independent, and their combinations are all the solutions.\n",
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