0
$\begingroup$

The differential equation $$\frac{d^nf(x)}{dx^n}=a.$$ My attempt I firstly solved the simpler version of it like $$\frac{df(x)}{dx}=a,$$$$\frac{d^2f(x)}{dx^2}=a,\ldots .$$ And got a generalistaion as $$ f(x)=\sum_{i=0}^n \frac{c_i x^i}{i!}$$ as $i$ goes from $0$ to $n$ where $c_i$ is constant (including a) but how can we generalize this with rigorous proof please help me I needed the solution if this differential equation to prove that in circular motion we can neglect the $n$th derivative of displacement as if we limit this result (if I am right) to infinity then it tends to zero.

$\endgroup$
4
  • 1
    $\begingroup$ You can prove it generally using induction. $\endgroup$
    – md2perpe
    Jul 2, 2020 at 15:37
  • 2
    $\begingroup$ You have a hypothesis about what the solution looks like for a given $n$. Now you only have to prove it, and that is probably best done using mathematical induction, i.e. showing it for a base case (e.g. $n=0$) and then showing that if it's true for $n=k$ then it is also true for $n=k+1.$ $\endgroup$
    – md2perpe
    Jul 2, 2020 at 15:40
  • $\begingroup$ MathJax works in the title section too, don't you know? $\endgroup$
    – Shaun
    Jul 2, 2020 at 15:52
  • $\begingroup$ Thank you md2perpe i prooved using induction Should i close my question or write the proof? $\endgroup$ Jul 2, 2020 at 16:04

1 Answer 1

1
$\begingroup$

Well, the solution is $$ f(x) = \sum_{k=0}^{n-1} \frac{ c_k x^k}{k!} + a \frac{x^n}{n!} $$ and differentiating this expression $n$ times show that indeed $$ \frac{d^n f(x)}{dx^n} = a $$ so it is a solution of the differential expression. You can use induction to compute the general derivative of a monomial $$\frac{d^n x^k}{dx^n}$$ and then use it in your proof.

$\endgroup$
1
  • $\begingroup$ Thank you triceratops i should have included in question that a is included in ci $\endgroup$ Jul 2, 2020 at 16:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .