Are $\sin$ and $\cos$ the only functions that satisfy the following relationship: $$ x'(t) = -y(t)$$ and $$ y'(t) = x(t) $$

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    $\begingroup$ Of course $\sinh$ and $\cosh$ are each other's derivatives without the sign difference, a situation which is easily understood by writing all four functions out in term of (potentially complex) exponentials. $\endgroup$ Jun 14, 2017 at 17:32
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    $\begingroup$ You could also open up more solutions by allowing more than two functions in the cycle. $\endgroup$
    – Brilliand
    Jun 14, 2017 at 20:03

5 Answers 5


The relationships $x'(t) = -y(t)$ and $y'(t) = x(t)$ imply $$x''(t) = -y'(t) = -x(t)$$ i.e. $$x''(t) = -x(t)$$ which only has solutions $x(t) = A \cos t + B \sin t$ for some constants $A$, $B$. For a given choice of the constants we then get $y(t) = -x'(t) = A \sin t - B \cos t$.


Basically, yes, they are. More precisely: if $x,y\colon\mathbb{R}\longrightarrow\mathbb{R}$ are differentiable functions such that $x'=-y$ and that $y'=x$, then there are numbers $k$ and $\omega$ such that$$(\forall t\in\mathbb{R}):x(t)=k\cos(t+\omega)\text{ and }y(t)=k\sin(t+\omega).$$


Your system is also satisfied by $$ x(t)=y(t)=0 $$

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    $\begingroup$ I don't see what utility these sort of obvious edge cases have. Given that it's clearly against the spirit of the question, I would think a comment is more than enough. $\endgroup$
    – Ant
    Jun 15, 2017 at 6:59
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    $\begingroup$ 'Edge' cases are important in mathematics. It's no more an 'edge' case than saying $2\sin$ and $2\cos$ are solutions. It's completely in the spirit of the question, it provides an example of a solution different to the trigonometric functions enquired about. @md2perpe +1 $\endgroup$
    – PM.
    Jun 15, 2017 at 8:06
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    $\begingroup$ Note that this is the same solution (A sin or cos t ) with A = 0, rather than an edge case $\endgroup$
    – Stilez
    Jun 15, 2017 at 8:57

We also have that relationship for $2\sin$ and $2\cos$

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    $\begingroup$ Well for all $k\sin x$ and $k\cos x$ $\endgroup$
    – Tim
    Jun 14, 2017 at 18:30

I suppose you can set: $$ x\left(t\right) = \exp\left(-i \cdot t\right)\\ y\left(t\right) = i\cdot\exp\left(-i \cdot t\right)\\ i = \sqrt{-1} $$


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