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?
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 $$y(t)=\frac{-kW\left(-\frac{e^{-\frac{v^2 (\text{C}+t)}{k}-1}}{k}\right)-k}{v}. $$