Linearity of y(t) = 2t²x(t) + 3t x(t − 4)

A system is linear if it obeys superposition: for inputs $x_1(t)\to y_1(t)$ and $x_2(t)\to y_2(t)$, the input $a x_1(t) + b x_2(t)$ must give $a y_1(t) + b y_2(t)$ for all constants $a$, $b$.

Given: $y(t) = 2t^2 x(t) + 3t\,x(t-4)$

Step 1: outputs for the individual inputs

$$\begin{aligned} y_1(t) &= 2t^2 x_1(t) + 3t\,x_1(t-4) \\ y_2(t) &= 2t^2 x_2(t) + 3t\,x_2(t-4) \end{aligned}$$

Step 2: output for the combined input $x_3(t) = a x_1(t) + b x_2(t)$

$$\begin{aligned} y_3(t) &= 2t^2\left[a x_1(t) + b x_2(t)\right] + 3t\left[a x_1(t-4) + b x_2(t-4)\right] \\ &= a\left[2t^2 x_1(t) + 3t\,x_1(t-4)\right] + b\left[2t^2 x_2(t) + 3t\,x_2(t-4)\right] \\ &= a\,y_1(t) + b\,y_2(t) \end{aligned}$$

Step 3: compare. The response to the weighted sum equals the weighted sum of responses, so superposition (additivity and homogeneity) holds. Also, zero input gives zero output.

The multipliers $2t^2$ and $3t$ depend on time, which makes the system time-varying, but they do not affect linearity, because the input appears only to the first power and there is no constant term.

Answer: $y(t) = 2t^2x(t) + 3t\,x(t-4)$ is a linear system (it is, however, time-variant).