Determine if the system described by the differential equation $$\ddot{y}(t) + 2\dot{y}(t)-6y(t) = 2x(t) $$ is linear, time-invariant and causal. $y(t)$ is the system output and $x(t)$ is the system input.
A system is linear if the input/output terms are linear - meaning there are no $y^2(t)$ or something like that. So the system is definitely linear.
A system is time-invariant if the coefficients of the differential equation are constants. So the system is definitely time-invariant.
A system is causal if the system's current output only depends on past and present inputs.
My question: Is it possible to determine from the differential if a system is causal/non-causal?
I used Mathematic to solve the system: $$ y(t) = k e^{\lambda_1 t} \left( c_1+ \int_1^t e^{-\lambda_1 \xi} x(\xi) d\xi \right) - ke^{\lambda_2 t} \left(c_2 + \int_1^t e^{-\lambda_2 \xi} x(\xi) d\xi\right), \tag{*} $$
where $\lambda_1 = -1+\sqrt{7}, \lambda_2 = -1-\sqrt{7}, k = \frac{1}{2 \sqrt{7}}$.
If $y(t)$ is bounded as $t\to \infty$ for any choice of $c_1,c_2$, then the system is causal, i.e. we can choose $c_1,c_2$ to meet the initial conditions and $y(t)$ only depends on the past values $x(s),s\leq t$. However, if at lest one of $c_1$ and $c_2$ has to be chosen so that the system does not "explode", then the future values of $x(s),s>t$ determine the current state. (Imagine it as preparing a dam for a future rain to prevent the dam braking when the rain season comes.)
Now looking at equation (*) we can see that the effect of $c_2$ diminishes over time as $\lambda_2<0$, but $c_1$ has to be put $$ c_1 = - \int_1^t e^{-a \xi} x(\xi) d\xi $$ so that $y(t)$ does not explode as $t\to \infty$, because $\lambda_1>0$.
In general, as system is causal when both of the roots $\lambda_1,\lambda_2$ of its characteristic equation have strictly negative real parts; and it is non-cause when at least one of them has strictly positive real part.