Lets extend to nth order derivative functions and not merely second order derivative functions.

The 4th order(say) derivative of the following functions is same the function itself

$$ \begin{aligned} f(x)&=e^{x}\\ f(x)&=0\\ f(x)&=\sin x\\ f(x)&=\cos x \end{aligned} $$ I was curious to know whether there are there any more such unique functions ?

  • 1
    $\begingroup$ Can you find the general solution of $\frac{d^4y}{dx^4}=y$? $\endgroup$ Aug 9, 2016 at 19:27

2 Answers 2


This type of problem is best approached via the characteristic equation. In this case, we have $y^{(4)} = y$, which yields characteristic equation $r^4 - 1 = 0$. The roots of this equation over $\mathbb{C}$ are $r = \pm 1, \pm i$, so the general solution to the differential equation is $$y = A_1e^{x} + A_2e^{-x} + A_3e^{ix} + A_4e^{-ix}, \quad A_i \in \mathbb{C}.$$ Using $\cos(x) = (e^{ix} + e^{-ix})/2$ and $\sin(x) = (e^{ix} - e^{-ix})/(2i)$ explains why your solutions arise for certain choices of the numbers $A_i$, and allows us to conclude that your list is almost complete. (You just forgot $e^{-x}$.) In other words, $e^{x}, e^{-x}, \cos(x), \sin(x)$ form a $\textit{basis}$ for the solution space.

  • $\begingroup$ More precisely, the $A_i$ are complex numbers and the family $(e^x,e^{-x},\sin x,\cos x)$ is a real basis of the space of real solutions. $\endgroup$
    – paf
    Aug 10, 2016 at 0:28
  • $\begingroup$ And a complex basis for the space of complex solutions, right? $\endgroup$
    – Mr. Chip
    Aug 10, 2016 at 0:30
  • $\begingroup$ Yes, that's right. $\endgroup$
    – paf
    Aug 10, 2016 at 0:32

Suppose you were looking for solutions to $\dfrac{d^ny}{dx^n}=y$. Obvious solutions might be:

  • $y_0 = 1 +\dfrac{x^{n}}{n!} +\dfrac{x^{2n}}{(2n)!}+\dfrac{x^{3n}}{(3n)!}+\cdots$
  • $y_1 = x +\dfrac{x^{n+1}}{(n+1)!} +\dfrac{x^{2n+1}}{(2n+1)!}+\dfrac{x^{3n+1}}{(3n+1)!}+\cdots$
  • $y_2 = \dfrac{x^{2}}{2!} +\dfrac{x^{n+2}}{(n+2)!} +\dfrac{x^{2n+2}}{(2n+2)!}+\dfrac{x^{3n+2}}{(3n+2)!}+\cdots$
  • $\cdots$
  • $y_{n-1} = \dfrac{x^{n-1}}{(n-1)!} +\dfrac{x^{2n-1}}{(2n-1)!}+\dfrac{x^{3n-1}}{(3n-1)!}+\dfrac{x^{4n-1}}{(4n-1)!}+\cdots$

and in fact these $\displaystyle y_j= \sum_{k=0}^{\infty} \dfrac{x^{kn+j}}{(kn+j)!}$ for $0 \le j \lt n$ and linear combinations of these $y_j$s are the only solutions to $\dfrac{d^ny}{dx^n}=y$ (including the constant $0$).

For example with $n=4$, you have solutions such as

  • $e^x=y_0+y_1+y_2+y_3$
  • $e^{-x}=y_0-y_1+y_2-y_3$
  • $\sin(x)=y_1-y_3$
  • $\cos(x)=y_0-y_2$
  • $\sinh(x)=y_1+y_3$
  • $\cosh(x)=y_0+y_2$

but you could also have something like $7y_1-2y_2$ as a solution


You must log in to answer this question.

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