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  "body": "\n## Proof\n\n1. Let $r \\in \\R$ and $e_r(x) = e^{r x}$. By the [[chain rule|0RXX2PT1MQ]] and [[the derivative of the exponential|0EX9GFJ0B3]], $e_r' = r\\, e_r$. By induction on $k$, $e_r^{(k)} = r^k e_r$ for each $k \\ge 0$.\n2. Write $L y = y^{(n)} + a_{n-1} y^{(n-1)} + \\dots + a_0 y$. By step 1, $L e_r = \\left(r^n + a_{n-1} r^{n-1} + \\dots + a_0\\right) e_r = p(r)\\, e_r$, where $p$ is the [[characteristic polynomial|0QDXMG46QW]].\n3. Since $p(r_k) = 0$, each $e_{r_k}$ is a solution of the [[homogeneous equation with constant coefficients|0A6PGTTP2A]] on $\\R$.\n4. Claim: if $s_1, \\dots, s_k$ are distinct reals, $c_1, \\dots, c_k \\in \\R$, and $\\sum_j c_j e^{s_j x} = 0$ for all $x \\in \\R$, then all $c_j = 0$. We argue by induction on $k$.\n5. For $k = 1$, $c_1 e^{s_1 x} = 0$ and [[$e^{s_1 x} \u003e 0$|0P9BNBW2W3]], so $c_1 = 0$.\n6. Let $k \\ge 2$ and assume the claim for $k - 1$ exponents. Multiply the relation by $e^{-s_k x}$. By the [[exponential of a sum|0E8HNVBCSP]], $\\sum_{j \u003c k} c_j e^{(s_j - s_k) x} + c_k = 0$ for all $x$.\n7. Differentiate, using step 1 and [[linearity of differentiation|0M0TKFRX2A]]: $\\sum_{j \u003c k} c_j (s_j - s_k)\\, e^{(s_j - s_k) x} = 0$ for all $x$.\n8. The numbers $s_j - s_k$ for $j \u003c k$ are distinct. By the induction hypothesis, $c_j (s_j - s_k) = 0$, and $s_j \\ne s_k$ gives $c_j = 0$ for $j \u003c k$.\n9. Then $c_k e^{s_k x} = 0$, so $c_k = 0$ by step 5. This proves the claim.\n10. By step 4, $e^{r_1 x}, \\dots, e^{r_n x}$ are [[linearly independent|0EVHJFTG9Z]] in the space $S$ of solutions on $\\R$.\n11. The constant coefficients are continuous on $\\R$. By [[the dimension theorem|0GK68PZ6KH]], $S$ has dimension $n$.\n12. By [[the basis criterion|096V7H13H9]], $n$ linearly independent vectors in $S$ form a basis. So $\\left(e^{r_1 x}, \\dots, e^{r_n x}\\right)$ is a [[fundamental system|0H4KNNCHPY]].\n13. Each solution is a combination of a basis, so it has the form $c_1 e^{r_1 x} + \\dots + c_n e^{r_n x}$. Conversely each such combination lies in the vector space $S$.\n\n## Idea\n\nEach root $r$ gives a solution $e^{r x}$, because differentiating $e^{r x}$ multiplies it by $r$. Exponentials with different rates cannot cancel: dividing by one of them and differentiating removes it, and induction does the rest. Then $n$ independent solutions fill the $n$-dimensional solution space.\n",
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