Can someone let me know linearity, time-invariance, causality, memory characteristic of the system? The input/output system is
$$
\frac{dy(t)}{dt}+2y(t)=2x^2(t)
$$



I want to know this is linear? causal? time-invariant? memoryless?

According to solution, the answer is 'Linear if zero initial conditions, causal, time-invariant, with memory'.



My trial:

$$y(t)=x^2(t)-\frac{1}{2}\frac{dy(t)}{dt}$$



Sorry for my poor drawing.




for input value: $x_1(t)=2x(t)$,
\begin{align}
y_1(t)&amp;=\left(2x(t)\right)^2+\left(-\frac12\right)\frac{d}{dt}\left(y(t)\right)\\
&amp;=4x^2(t)-\frac12\frac{dy(t)}{dt}\\
2y(t)&amp;=2x^2(t)-\frac{dy(t)}{dt}\\
\therefore y_1(t) &amp;\ne 2y(t)
\end{align}

$\because$ not satisfying Homogeneity, Non-Linear.





The system doesn't have $x(t+k)$. (It means $y(t)$ is not made by $x(t+k)$)


$\therefore$ The system is causal because




for input value: $x_1(t)=x(t-t_0)$,
\begin{align}
y_1(t)&amp;=x^2(t-t_0)-\frac{1}{2}\frac{dy(t)}{dt}\\
y(t-t_0)&amp;=x^2(t-t_0)-\frac{1}{2}\frac{dy(t-t_0)}{dt}\\
\therefore y_1(t) &amp;\ne y(t-t_0)
\end{align}

$\therefore$ time-varying.
The system doesn't have any $\displaystyle\int$.

i.e.) $\displaystyle\int x(\tau) d\tau$.

$\therefore$ memoryless




What is the real solution?
There's an error in your 3rd section (linearity) - you replace $y(t)$ with $y_1(t)$ on the left side of the equation but not on the right side. You should have
$$y_1(t)=x^2(t-t_0)-\frac{1}{2}\frac{dy_1(t)}{dt}$$.
Now you see that $y_1(t)=y(t-t_0)$.

As for memoryless - start with your initial equation and integrate both sides. Now you have
$$
\int y(t)dt=\int x^2(t)dt-\frac{1}{2} y(t)dt
$$

and rearranging gives
$$
y(t) = 2 \int x^2(t)dt-2 \int y(t)dt
$$
Does that help?