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How to solve this differential equation:

$\frac{d^nf(x)}{dx^n}=\pm k^2f(x)$

For $n=1,2,3$ and $\forall n\in\mathbb{N}$, and both signs, if this is possible.

I encounter these often in physics, with solutions but no derivations. I would like to know how they are solved.

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    It is a linear homogeneous differential equation with constant coefficients. A differential equations textbook should explain the method to find the general solution. In fact, lots of things "encountered in physics with no derivations" can be found in mathematics textbooks. For courses which physics departments no longer require, in favor of physics courses with no derivations...2012-07-18

2 Answers 2

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This equation is linear in $f(x)$, so if you just find $n$ linearly independent solutions by wild-guessing it should be enough to prove that this is the general solution of this equation.

For instance, since you know easily that for $n = 1$ you need to substitute $f(x) = e^{rx}$, a wild guess would be the same guess, which leads to $ r^n e^{rx} = \pm k^2 e^{rx} \quad \Longrightarrow \quad r^n = \pm k^2 \quad \Longrightarrow \quad r = \sqrt[n]{\pm k^2} e^{\frac{2\pi i j}n}, \quad 0 \le j \le n-1. $ Since all those values of $r$ are distinct, you have $n$ solutions (for a fixed $n^{\text{th}}$ root of $+k^2$ or $-k^2$). Tadam!

Hope that helps,

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Assume a solution

$ f\left(x\right) \propto \exp\left(\lambda x\right), $ which gives $ \lambda^n = \pm k^2. $ Now let $ \lambda = k^{2/n} \exp\left(i r\right), $ so that $ \lambda^n = k^2 \exp\left(i n r\right) = \pm k^2 \Rightarrow \exp\left(i n r\right) = \pm 1. $ If $+1$, $ r_m = \frac{2 \pi}{n} m, $ where $m = 0, 1, ..., n - 1$.

If $-1$, $ r_m = \frac{\pi}{n} \left(2m+1\right), $ where $m = 0, 1, ..., n - 1$.

Your solution is then $ f\left(x\right) = \sum_{m=0}^{n-1} a_m \exp\left[k^{2/n} \exp\left(i r_m\right) x\right], $ where $a_m$'s are determined with $n$ initial conditions and the correct $r_m$'s are used based on whether it is $+1$ or $-1$.

You can check this for the simple harmonic oscillator ($n = 2$, and the $-1$ case).