# Solution to a general-derivative ODE $f^{(n)}(x)=f(x)$?

Is there a way to find / write a solution to a general from ODE $$f^{(n)}(x)=f(x)$$ ? For a particular $$n$$ it is easy, see an example for instance for $$n=7$$ below from Wolfram|Alpha. So it is obviously algorithmic. Any sources, reference, ideas, how to write it for a general $$n$$ explicitly? Thank you!

https://www.wolframalpha.com/input/?i=f%28x%29%3D%3DD%5Bf%28x%29%2C%7Bx%2C7%7D%5D

• Consider $f(x) = Ce^{\lambda x}$. You'll get a condition on $\lambda$. Apr 10 '21 at 18:43
• ..and if you found $n$ linearly independent solutions, Bob's your uncle, since this is the dimension of the solution vector space.
– user145413
Apr 10 '21 at 18:51
As suggested by @Cameron Williams in the comments - consider the ansatz $${f(x) = Ce^{\lambda x}}$$. Then $$\Rightarrow C\lambda^n e^{\lambda x} = Ce^{\lambda x}$$ this of course implies that $${\lambda^n = 1}$$. There will be $${n}$$ distinct $${\lambda}$$ satisfying $${\lambda^n}$$. To see this, note $$\lambda^n = 1 \Rightarrow \lambda^n = e^{2\pi i k }\ | k \in \mathbb{Z}\Rightarrow \lambda = e^{\frac{2\pi i k}{n}}\ |\ k \in \{0,1,\dots ,k-1\}$$ I will label the solutions as $${\lambda_i\ |\ i \in \{0,\dots,k-1\}}$$ for simplicity. Thus $${f_i(x) = C_ie^{\lambda_i x}}$$ are a set of $$n$$ linearly independent solutions to the ODE, and hence we can be sure that is all of them by what @username said also. The ODE is linear, and so the most general solution is given by the sum: $$f(x) = C_0e^{\lambda_{0} x} + \dots + C_{k-1}e^{\lambda_{k-1} x} = C_1e^{\lambda_1 x} + \dots + C_{k-1}e^{\lambda_{k-1} x } + C_{0}$$ (since $${\lambda_0 = 1}$$). As desired.