I'm a high school senior currently taking AP Calculus AB. I'm trying to learn a little about dynamical systems in my free time (yes, I know the pre-requisites are supposed to be multi-variable calculus and linear algebra, but I was eager!)

Basically, I'm starting Strogatz's Nonlinear Dynamics and Chaos and came across the equation $\dot x = \sin x$, which I presume means the same thing as $x'(t)=\sin(x(t))$ since my understanding is that $x$ is a function of $t$.

The separation of variables seems pretty straightforward -- you get $t = -\ln|\csc x + \cot x| + C$.

I agree with Strogatz on that much. But then I wondered -- wasn't a solution to a differential equation supposed to be a function $x(t)$, not a variable $t$? What the heck does it even mean to have a solution expressed as $t = ...$ ? Indeed, when I told Wolfram Alpha to solve $x'(t) = \sin(x(t))$, it returned $x(t) = 2\cot^{-1}(e^{c_1 - t})$ -- quite different from the solution given above, although I have no idea how Wolfram Alpha got it or how their process differred from Strogatz's.

Below is an image of the Strogatz page in question (section 2.1). Having not understood why solving for $t$ was useful, I am of course completely (even more) lost at the part where he says $x = x_0$ and does some stuff with initial conditions or whatever.

Basically, isn't the point of solving a differential equation generally to find the function $x(t)$ that satisfies it? So why isn't Strogatz doing that here, and what else is it that he's doing instead?

enter image description here

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    $\begingroup$ It is called implicit solution. If you want you could solve for $x$ and everything's fine. $\endgroup$
    – Azlif
    Nov 21, 2019 at 1:51

1 Answer 1


The WA answer is correct although not the simplier

$$x'=\sin(x)$$ $$ \implies \int \frac {dx}{\sin x}=\int dt$$ Using tangent half angle substitution $$dx =\frac {2dw}{1+w²}, \sin x =\frac {2w}{1+w²}$$ $$ \int \frac {dw}w=t+k$$ $$\ln w =t+k \implies w=ke^t$$ Susbtitute back $w=\tan (x/2)$ $$\tan (x/2)=ke^t$$ That you can also write as WA $$\cot (x/2)=ke^{-t}$$ $$ \implies x=2\cot^{-1}(ke^{-t})$$ Note that $\dot {x}$ is Newton's notation for derivative. Where $x'$ is Lagrange's notation.

For Strogatz's solutiuon, you have to note that $$ E=-\ln(\csc x+\cot x)$$ $$ E=-\ln\left(\frac 1 {\sin(x)}+\frac {\cos x}{\sin x}\right)$$ Use trigonometric substitution $t=\tan (x/2), \sin x= \frac {2t}{1+t^2}, \cos x =\frac {1-t^2}{1+t^2}$ $$E=-\ln \left(\frac 1 t \right)$$ $$E=\ln t \implies E=\ln(\tan(x/2))$$ Stoglatz keep the implicit notation as it's more simple. But WA answer and Stoglatz's answers are the same, they don't differ.

  • $\begingroup$ What does $\frac {dx}{\sin x}$ mean? Are you essentially "multiplying" $dx$ by $\csc x$? And, if so, is it generally true that you can treat $dx$ as if it's being multiplied by everything else in an integral? $\endgroup$
    – Will
    Nov 21, 2019 at 3:36
  • $\begingroup$ Also, was I correct to say that $x = x(t)$ in this problem, so that $x'$ means the same thing as $x'(t)$? Wolfram Alpha uses the $x(t)$ notation, which is different from yours.... $\endgroup$
    – Will
    Nov 21, 2019 at 3:37
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    $\begingroup$ yes that's correct. And at $t=0 ,x=x_0$ means $x(0)=x_0$ @Will $\endgroup$ Nov 21, 2019 at 3:44
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    $\begingroup$ you integrate both side. it has meaning within the integrand. $\endgroup$ Nov 21, 2019 at 3:46
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    $\begingroup$ I am not sure to undertsand your question @Will. We integrate both side of the equation. $\endgroup$ Nov 21, 2019 at 3:49

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