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Suppose we have 2 variables $x$ and $y$ and their relationship is:

Suppose that we have another equation such as $y = dx/dt$ and we want to solve that nonlinear system of equations.

The time constant of a system is the time interval it takes 5 times of it to go from the starting value to the final value.

My first idea of finding the time constant of that set of equations($y = dx/dt$ and the picture) above is to go through the non-linear graph and imagine that for infinitesmall changes it becomes linear so that then we could find infinite time constants of infinite small intervals in the xy graph then add them up but that process is $\int^{\mathrm{curve}}(\mathrm{curve})'dx = [\mathrm{curve}]_{\mathrm{start}}^{\mathrm{final}} $ and thats how you get up a expectation value for the average of the curve regardless of if the xy graph of the picture was linear or not,so the time constant of the system described by the set of equations is $[\frac{x}{y}]_{\mathrm{start}}^{\mathrm{final}}$. So regardless of if a system is linear or not we can know at some times what value that system would have. Am I correct on my assumptions?

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  • $\begingroup$ I'm unsure if I understand your question, but from what I gather, your "time constant", 5T ~ final value rule belongs to linear first-order dynamics. For a general and nonlinear f(x) there is no single, global "time constant" (where f(x) = y = dx/dt ->dt = dx/f(x)) So, if you only know the graph numerically, compute the time by integrating the tabulated function 1/f(x), meaning t(x0->x1) = integral^x1_x0 (dx/f(x)) $\endgroup$ Commented Aug 21 at 18:48

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What you have is the ODE $y=f(x)=\dfrac{dx}{dt}$. It is of the separable type and you solve it as

$$\int_{x_s}^{x_f}\frac{dx}{f(x)}=t_f-f_s.$$

If the the starting and final values are known in terms of $y$, then

$$\int_{f^{-1}(y_s)}^{f^{-1}(y_f)}\frac{dx}{f(x)}=t_f-f_s.$$

In your example, we can assume that $y=100\,2^x$, giving

$$\dfrac{1}{100\log(2)}(2^{-x_s}-2^{-x_f})=\dfrac{1}{\log(2)}\left(\dfrac1{y_s}-\dfrac1{y_f}\right)=t_f-t_s.$$

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No. The [x/y] idea is not generally correct. You have an autonomous first order ODE, dx/dt = y = f(x)

As mentioned, separate your variables and compute time by integrating.

$\frac{dx}{dt} = f(x) \Rightarrow dt = \frac{dx}{f(x)} \ \ \Big\vert \ \int(...) \ \Rightarrow t(x_0\rightarrow x_1) = \int\limits^{x_1}_{x_0}\frac{dx}{f(x)} $

(indeed, d/dx (x/y) is not equal to 1/y in general, because here y depends on x and the extra term after differentiation is generally nonzero)

Useful notions to take into account

The 5τ rule applies only when the motion is (approximately) exponential near the final value.

let's look into it, $\ \frac{dx}{dt}= -\frac{1}{\tau}(x-x_\infty) \Rightarrow x(t)=x_\infty \ + \ (x_0-x_\infty)e^{\frac{-t}{\tau}}$

Now if we plug 5τ in, we're getting only close at about 99% of $x_\infty$, taking $ 1-e^{-5} $ as the fraction reached.

Remember, you are not working with first order linear dynamics

Depending on f(x), strange things can happen if f(x) grows really fast with x. One single time constant cannot possibly capture that.

Also, if $\ x_{eq}\ $ satisfies $\ f(x_{eq}) = 0, \ \ f'(x_{eq})<0,\ $ then linearize $f(x) \approx f'(x_{eq})(x-x_{eq})$

Then I believe you may speak of a local time constant $\ \tau_{eq} = -\frac{1}{f'(x_{eq})}$

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