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Let $y_0$ be a solution to the ODE: $$y^{(n)} + a_{n-1} y^{(n-1)} + \dots + a_0 y = 0$$

In this note (part 2), it is claimed that $y_0$ must be infinitely differentiable because we can write $y^{(n)}$ in terms of derivatives with lower order.

My understanding is that since $y^{(n)} = -a_{n-1} y^{(n-1)} - \dots - a_0 y$, differentiating both sides will give you a formula of $y^{(n + 1)}$ in terms of derivatives with lower order. Continuing this process gives us formula for derivative of arbitrarily high order. My problem with this is that why are allowed to differentiate on both sides? Aren't we allow to differentiate only if $y^{(n)}$ is differentiable?

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Yes, but $y^{(n)}$ is differentiable because

  1. $y$ is $n$ times differentiable, since it is a solution of the differential equation.
  2. $y^{(n)} = -a_{n-1}\,y^{(n-1)} - \dots - a_0\, y$, and each $y^{(k)}$, $0\le k<n$, is differentiable.
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You have the well-known regularity result: $$\text{if }f\in \mathcal{C}^k \text{ then each solution of } y^{(n)}=f(t,y(t),...,y^{(n-1)})\text{ is } \mathcal{C}^{n+k}.$$ In the case of the homogenous ODE with constant coefficient, we have $$f(t,u_1,...,u_n)=-a_{n-1} u_n - \dots - a_0 u_1,$$ which is infinitely differentiable on $\mathbb{R}\times\mathbb{R}^n$; so $y$ is infinitely differentiable on $\mathbb{R}$.

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