How can we solve this differential equation $d^nf(x)/dx^n=a$?

Question

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.

Answer 1

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.