1
$\begingroup$

I have arrived at the equation below to describe the dynamics of a system. As you can see, state variable $y$ is equal to an integral of itself.

$$y(t)=k\int_0^t\frac{1}{y(\tau)}d\tau$$

I'd like to derive an analytical solution to this. My initial thought is to differentiate this equation wrt $t$, which gives (I think):

$$\frac{dy}{dt}= \frac{k}{y}$$

Then, I rearrange and integrate to give:

$$\int y~dy = \int k~dt$$

$$y = \sqrt{2kt}+c$$

I think this is correct.

Now, say I have another term in my equation, like so:

$$y(t)=k\int_0^t\frac{1}{y(\tau)}d\tau + vt$$

If I adopt a similar approach, I arrive at the following:

$$\frac{y^2}{2}=\int(k+vy)~ dt$$ $$\frac{y^2}{2}=kt + \int vy ~dt$$

It's the second term on the RHS that I'm not sure what to do with, given that $y(t)$. How do I solve this?

$\endgroup$
0
1
$\begingroup$

Your approach to the first problem is totally what I would do. Not sure about the second one: \begin{align*} y(t)&=k\int_0^t\frac{1}{y(\tau)}d\tau + vt \\ \dot{y}(t)&=\frac{k}{y(t)}+v =\frac{k+vy}{y} \\ \int\frac{y\,dy}{k+vy}&=\int dt \\ \frac{y}{v}-\frac{k \log (k+v y)}{v^2}&=t+C. \end{align*} You can solve this for $y$ using the Lambert $W$ function. The result is $$y(t)=\frac{-kW\left(-\frac{e^{-\frac{v^2 (\text{C}+t)}{k}-1}}{k}\right)-k}{v}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.