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  "body": "\n## Proof\n\nWe factor the operator through $(D - r)^m$.\n\n1. Let $D$ be differentiation on the infinitely differentiable functions $\\R \\to \\R$. For a real polynomial $q(t) = \\sum_i b_i t^i$, let $q(D) u = \\sum_i b_i u^{(i)}$. The equation is $p(D) y = 0$, with $p$ the [[characteristic polynomial|0QDXMG46QW]].\n2. For real polynomials $q_1, q_2$, we have $q_1(D)\\left(q_2(D) u\\right) = (q_1 q_2)(D) u$. Indeed, $D^i$ is linear by [[linearity of differentiation|0M0TKFRX2A]] and $D^i D^j = D^{i+j}$. Expanding both sides gives the same sum.\n3. Let $s \\in \\R$ and $k \\ge 0$. By the [[product rule|0V9A43H7WV]], the [[chain rule|0RXX2PT1MQ]] and [[the derivative of the exponential|0EX9GFJ0B3]], $D\\left(x^k e^{s x}\\right) = k x^{k-1} e^{s x} + s\\, x^k e^{s x}$. So $(D - s)\\left(x^k e^{s x}\\right) = k x^{k-1} e^{s x}$, which is $0$ for $k = 0$.\n4. Applying step 3 $j$ times, $(D - s)^j \\left(x^k e^{s x}\\right)$ is a constant multiple of $x^{k-j} e^{s x}$ for $j \\le k$. It is $0$ for $j = k + 1$, and so for each $j \u003e k$.\n5. Solutions. Since $r$ has [[multiplicity|03XVP8Q4GY]] $m$, $p(t) = h(t) (t - r)^m$ for a real polynomial $h$. Let $0 \\le k \\le m - 1$.\n6. By steps 2 and 4, $p(D)\\left(x^k e^{r x}\\right) = h(D)\\left((D - r)^m \\left(x^k e^{r x}\\right)\\right) = h(D)\\, 0 = 0$. So $x^k e^{r x}$ solves the [[equation with constant coefficients|0A6PGTTP2A]] on $\\R$.\n7. Lemma. Let $s_1, \\dots, s_k$ be distinct reals and $q_1, \\dots, q_k$ real polynomials with $\\sum_j q_j(x) e^{s_j x} = 0$ for all $x \\in \\R$. Then each $q_j$ is the zero polynomial. We argue by induction on $k$.\n8. For $k = 1$, [[$e^{s_1 x} \u003e 0$|0P9BNBW2W3]], so $q_1(x) = 0$ for every $x \\in \\R$. A nonzero polynomial [[has finitely many roots|0KJVQV3E0F]], so $q_1 = 0$.\n9. Let $k \\ge 2$. Multiply by $e^{-s_k x}$ and use the [[exponential of a sum|0E8HNVBCSP]]: $\\sum_{j \u003c k} q_j(x) e^{\\mu_j x} + q_k(x) = 0$, where $\\mu_j = s_j - s_k \\ne 0$ are distinct.\n10. For a polynomial $q$ and $\\mu \\ne 0$, step 3 gives $D\\left(q(x) e^{\\mu x}\\right) = \\left(q' + \\mu q\\right)(x)\\, e^{\\mu x}$. If $q \\ne 0$ has degree $d$ and leading coefficient $a$, then $q' + \\mu q$ has degree $d$ and leading coefficient $\\mu a \\ne 0$, by the [[derivative of a polynomial|0MH1WK89PP]].\n11. Let $e$ be one more than the degree of $q_k$, or $e = 0$ if $q_k = 0$. Differentiating the relation of step 9 $e$ times kills $q_k$. By step 10, it gives $\\sum_{j \u003c k} \\tilde q_j(x) e^{\\mu_j x} = 0$ with $\\tilde q_j \\ne 0$ whenever $q_j \\ne 0$.\n12. By the induction hypothesis, each $\\tilde q_j = 0$, so each $q_j = 0$ for $j \u003c k$. Then $q_k(x) e^{s_k x} = 0$, and $q_k = 0$ by step 8.\n13. Independence. If $\\sum_{k=0}^{m-1} c_k x^k e^{r x} = 0$ on $\\R$, the lemma with one exponent gives $\\sum_k c_k x^k = 0$ as a polynomial. So each $c_k = 0$, and the $m$ functions are [[linearly independent|0EVHJFTG9Z]].\n14. Now assume all roots of $p$ are real. We show $p(t) = \\prod_{i=1}^{n} (t - c_i)$ with real $c_i$, by induction on $n$. For $n = 0$ the product is empty.\n15. For $n \\ge 1$, the [[fundamental theorem of algebra|0JP0W07Y9A]] gives a root $c$ of $p$, real by assumption. By the [[factor theorem|0K3MWA7DMC]] in $\\R[t]$, $p = (t - c)\\, p_1$ with $p_1$ real and monic of degree $n - 1$.\n16. Each complex root $z$ of $p_1$ is a root of $p$, since $p(z) = (z - c) p_1(z)$. So it is real, and the induction hypothesis applies to $p_1$.\n17. Let $\\rho_1, \\dots, \\rho_s$ be the distinct values among the $c_i$, with $\\rho_j$ occurring $m_j$ times. Then $p(t) = \\prod_j (t - \\rho_j)^{m_j}$ and $m_1 + \\dots + m_s = n$. Each root $z$ of $p$ makes some factor vanish, so the roots are exactly $\\rho_1, \\dots, \\rho_s$.\n18. The multiplicity of $\\rho_j$ is $m_j$. It is at least $m_j$ by step 17. Suppose $(t - \\rho_j)^{m_j + 1}$ divides $p$, say $p = (t - \\rho_j)^{m_j + 1} g$.\n19. Since [[$\\R[t]$ is an integral domain|0B5Z1EC66J]], we may [[cancel|0GJM24EXP9]] $(t - \\rho_j)^{m_j}$. This gives $\\prod_{i \\ne j} (t - \\rho_i)^{m_i} = (t - \\rho_j)\\, g$. At $t = \\rho_j$ the left side is nonzero and the right side is $0$, a contradiction.\n20. So the families for the distinct roots have $m_1 + \\dots + m_s = n$ functions in total, all solutions by step 6.\n21. They are linearly independent. A vanishing combination can be grouped as $\\sum_j q_j(x) e^{\\rho_j x} = 0$, where the coefficients of $q_j$ are the coefficients used for $\\rho_j$. By the lemma each $q_j = 0$, so all coefficients vanish.\n22. The space $S$ of solutions on $\\R$ has dimension $n$, by [[the dimension theorem|0GK68PZ6KH]]. By [[the basis criterion|096V7H13H9]], these $n$ independent solutions form a basis, that is, a [[fundamental system|0H4KNNCHPY]].\n\n## Idea\n\nThe operator $p(D)$ factors like the polynomial $p$, and the factor $(D - r)$ lowers the power of $x$ in $x^k e^{r x}$ by one. So $(D - r)^m$ kills $x^k e^{r x}$ for $k \u003c m$. Independence follows because differentiating a polynomial times an exponential keeps the degree, while differentiating a polynomial alone lowers it.\n",
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