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What kind of ODE is the following one:

$y^{(n)}(t) = \dfrac{x^{(n)}(t)}{f(n)}$

I have never seen such type. Any references would be very much appreciated. Thanks

Mr. N
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    If $y^{(n)}=\frac{x^{(n)}}{f(n)}$, then $y=\frac{x}{f(n)}+P$ where $P$ is a polynomial of degree at most $n-1$ that can be computed with some additional initial conditions. – Tuvasbien Dec 30 '20 at 03:12
  • Just that simple?! Thanks! – Mr. N Dec 30 '20 at 03:14
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    I know this might be a bit off topic but this ODE somehow resembles the $n^{th}$ derivative of $e^{x^{2}}$ – B E I R U T Dec 30 '20 at 03:15
  • May I ask you, @GAUSS1860, to explain and show the resemblance? – Mr. N Dec 30 '20 at 03:17
  • https://math.stackexchange.com/questions/193702/find-an-expression-for-the-%24n%24-th-derivative-of-%24f%28x%29%3De%5E%7Bx%5E2%7D%24#193730 Refer to Thomas Andrews answer. – B E I R U T Dec 30 '20 at 03:19

1 Answers1

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If

$$ y^{(n)}(t) = \dfrac{x^{(n)}(t)}{f(n)} $$

then we have

$$ y(t) = \dfrac{x(t)}{f(0)},\quad y^\prime(t) = \dfrac{x^\prime(t)}{f(1)},\quad y^{\prime\prime}(t) = \dfrac{x^{\prime\prime}(t)}{f(2)},\text{ etc.} $$

from which it follows that $f(0)=f(1)=f(2),\cdots$.

So, essentially you have $$ y(t)=c\cdot x(t) $$

John Wayland Bales
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