When I want to prove something in general about all differentiable functions from $f:\mathbb{R}\to\mathbb{R}$ where $f'(x)=f(x)$ or $f''''(x)=f(x)$, how do I find them, and how do I know I've found them all?

I can give examples, $f(x)=0$ and $f(x)=e^x$ and $f(x)=\sin(x)$ and $f(x)=\cos(x)$

I can find other examples, but I'm not sure how to guarantee I've found them all.

  • $\begingroup$ for $f:\mathbb C\to\mathbb C$, there is one linearly independent solution to $f'(x)=f(x)$, and there are four to $f''''(x)=f(x)$ $\endgroup$ Jun 29, 2021 at 1:16
  • $\begingroup$ Your “solutions” indicate you want $f’(x)$ or $f’’’’(x)=f(x),$ not that $f’(x)=f’’’’(x),$ as your question implies. $\endgroup$ Jun 29, 2021 at 1:20
  • $\begingroup$ In any even, any solution to $f(x)=f’(x)$ is a solution to $f(x)=f’’’’(x).$ So you don’t need the two separate equations. $\endgroup$ Jun 29, 2021 at 1:23
  • $\begingroup$ The degree of the derivative isnt getting at what I wanted to know. My question is more how so we find solution to $y^n = y$ for $n \in \mathbb{Z}$ like @user0102 said below which is a question I can answer $\endgroup$ Jun 29, 2021 at 1:28

2 Answers 2


You can solve it through separation of variables or through the integrating factor method.

Let us stick with the second:

\begin{align*} y' = y & \Longleftrightarrow y' - y = 0\\\\ & \Longleftrightarrow \exp(-x)y' - \exp(-x)y = 0\\\\ & \Longleftrightarrow [\exp(-x)y]' = 0\\\\ & \Longleftrightarrow \exp(-x)y = k\\\\ & \Longleftrightarrow y = k\exp(x) \end{align*}

where $k\in\mathbb{R}$ and $x\in\mathbb{R}$.

Hopefully this helps!

  • $\begingroup$ Why is it clear, in a larger $k$ scenario, that this produces all the solutions of the equation? In other words, what criteria of uniqueness would apply? Since such equations aren't necessarily "first order", they wouldn't apply. You could, however, convert this equation into a first order equation by using vector-valued functions. Then I think the equation is linear and "first order" with respect to the constraints, and therefore admits a unique solution. $\endgroup$ Jun 29, 2021 at 1:23

It is known that the vector space of the solutions to $y^{(n)}=y$ over the complex numbers is generated by $e^{\zeta^kx}$ where $\zeta=e^{2\pi i/n}$ and $k\in\mathbb Z/n\mathbb Z$. Thus, we are looking for complex numbers $c_k$ such that $f(x)=\sum_{k\in\mathbb Z/n\mathbb Z}c_ke^{\zeta^kx}\in\mathbb R$ for each $x\in\mathbb R$.

For each $x\in\mathbb R$, we need $\overline{f(x)}=f(x)$, or $$\overline{f(x)}=\sum_{k\in\mathbb Z/n\mathbb Z}\overline{c_k}e^{\zeta^{-k}x}=\sum_{k\in\mathbb Z/n\mathbb Z}c_ke^{\zeta^kx}=f(x).$$ Thus, we need $\overline{c_k}=c_{-k}$, and the solution space is generated by

$$\left\{\frac{e^{\zeta^kx}+e^{\zeta^{-k}x}}2=e^{\cos(\frac{2\pi k}{n})x}\cos\left(\sin\!\big(\frac{2\pi k}n\big)x\right),\frac{e^{\zeta^kx}-e^{\zeta^{-k}x}}{2i}=e^{\cos(\frac{2\pi k}{n})x}\sin\left(\sin\!\big(\frac{2\pi k}n\big)x\right):k\in\mathbb Z/n\mathbb Z\right\}$$ over $\mathbb R$.


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