Causality as applied to capacitors This question stems from a point of confusion that I still have about the causality, linearity, and time-invariance in LCCDEs. I wanted to use the capacitor as an example. 

Consider a capacitor with capacitance $C$. Taking the current $i(t)$ to be the input to the system and the voltage $v(t)$ to be the output we have $$i(t) = C \frac{\mathrm{d} v(t)}{\mathrm{d}t}$$

This differential equation can be solved to obtain
 $$v(t) = v(t_0) + \frac{1}{C} \int_{t_0}^{t} i(\tau) \mathrm{d} \tau$$

My first question is: isn't this mathematically valid for all $t$? In other words, does this give us the response for all $t$ or is it only valid for $t &gt; t_0$? If it is valid for all $t$, including the $t &lt; t_0$ case, doesn't this make the system non-causal since it anticipates future input and output values? Are we allowed to integrate backwards in time?

My second question relates to the assertion that a for the LCCDE to describe a linear system, the initial conditions must be zero. Suppose $t_0 = 0$ such that  $$v(t) = v(0) + \frac{1}{C} \int_{0}^{t} i(\tau) \mathrm{d} \tau$$

With $v(0) = 0$ the system is linear. But the choice of $t_0 = 0$ is arbitrary, since for example
$$v(t) = v(0) + \frac{1}{C} \int_{0}^{t} i(\tau) \mathrm{d} \tau = v(2) + \frac{1}{C} \int_{2}^{t} i(\tau) \mathrm{d} \tau$$

Why shouldn't we require that $v(2) = 0$ as well for that matter? What am I missing here? Thank you in advance.
Your first equation is indeed valid for all $t$, but this has nothing to do with causality. You have

$$\begin{align}v(t)&amp;=\frac{1}{C}\int_{-\infty}^ti(\tau)d\tau\\&amp;=\frac{1}{C}\int_{-\infty}^{t_0}i(\tau)d\tau+\frac{1}{C}\int_{t_0}^ti(\tau)d\tau\\&amp;=\frac{1}{C}\int_{-\infty}^{t_0}i(\tau)d\tau-\frac{1}{C}\int_{t}^{t_0}i(\tau)d\tau\end{align}$$

If $t_0&gt;t$ you simply subtract from the first integral the contribution from the interval $[t,t_0]$. But this is still causal, you don't need the future input, you just need to subtract it in case you added it in the first integral. The output could of course be computed without it.

Concerning your second question, you choose the time $t_0$ as the beginning of your processing. The value of $t_0$ doesn't matter but once you choose it, it's fixed. The system is linear if the initial conditions at $t=t_0$ are zero.