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!


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    $\begingroup$ Consider $f(x) = Ce^{\lambda x} $. You'll get a condition on $\lambda $. $\endgroup$ Apr 10 '21 at 18:43
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    $\begingroup$ ..and if you found $n$ linearly independent solutions, Bob's your uncle, since this is the dimension of the solution vector space. $\endgroup$
    – user145413
    Apr 10 '21 at 18:51
  • $\begingroup$ @CameronWilliams and username - I have written an answer based on your comments $\endgroup$ Apr 10 '21 at 20:01

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.


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