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  "body": "\n## Proof\n\n1. Let $u = c_1 y_1 + c_2 y_2$. As solutions of the [[homogeneous linear equation|0EQZSAB9EC]], $y_1$ and $y_2$ are $n$ times differentiable on $I$, with $L y_1 = 0$ and $L y_2 = 0$.\n2. We show by induction on $k$ that, for $0 \\le k \\le n$, $u$ is $k$ times differentiable on $I$ and $u^{(k)} = c_1 y_1^{(k)} + c_2 y_2^{(k)}$.\n3. For $k = 0$ this is the definition of $u$.\n4. Let $k \u003c n$ and assume the claim for $k$. The functions $y_1^{(k)}$ and $y_2^{(k)}$ are differentiable, since $k \u003c n$.\n5. By [[linearity of differentiation|0M0TKFRX2A]], $u^{(k)} = c_1 y_1^{(k)} + c_2 y_2^{(k)}$ is differentiable with derivative $c_1 y_1^{(k+1)} + c_2 y_2^{(k+1)}$. This proves the claim for $k + 1$.\n6. Write $a_n = 1$. By step 2, at each $x \\in I$, $L u = \\sum_{k=0}^{n} a_k \\left(c_1 y_1^{(k)} + c_2 y_2^{(k)}\\right) = c_1 L y_1 + c_2 L y_2$.\n7. By step 1, this is $c_1 \\cdot 0 + c_2 \\cdot 0 = 0$. So $u$ is $n$ times differentiable and solves the [[linear equation|0PB7K8M7FG]] $L y = 0$ on $I$.\n\n## Idea\n\nDifferentiation is linear, and $L$ is built from derivatives by multiplying with functions and adding. So $L$ is linear, and it maps a combination of solutions to the same combination of zeros.\n",
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