Among the following, the differential equation(s) not representative of a linear dynamical system is/are ____. (i) \[ \frac{d^2x}{dt^2} + 2t\frac{dx}{dt} + x(t) = u(t) \] (ii) \[ \left(\frac{dx}{dt}\right)^2 + 2\frac{dx}{dt} + x(t) = u(t) \] (iii) \[ \frac{d^2x}{dt^2} + 2\frac{dx}{dt} + x(t) = u(t) \]
Show Hint
Linearity only cares about whether \(x\), \(u\), and their derivatives appear to the first power with no products between them; time-varying coefficients alone do not break linearity.
Step 1: Understanding the Concept.
A clean way to test linearity without memorizing a rule is to apply the superposition test directly: pick two arbitrary solutions $x_1(t)$ and $x_2(t)$ (produced by inputs $u_1(t)$ and $u_2(t)$), form the combination $x = \alpha x_1 + \beta x_2$, and check whether it satisfies the same equation with input $\alpha u_1 + \beta u_2$. If it does for every term, the equation is linear.
Step 2: Key Formula or Approach.
For any term built only from $x$ and its derivatives raised to the first power, superposition works automatically because differentiation is a linear operation: $\frac{d}{dt}(\alpha x_1 + \beta x_2) = \alpha \frac{dx_1}{dt} + \beta \frac{dx_2}{dt}$. Superposition breaks the moment a term involves $x$ or a derivative of $x$ multiplied by itself or by another such term, since $(\alpha a + \beta b)^2 \neq \alpha a^2 + \beta b^2$ in general.
Step 3: Detailed Explanation.
Test equation (i): every term, $\frac{d^2x}{dt^2}$, $2t\frac{dx}{dt}$, and $x(t)$, is $x$ or a derivative of $x$ multiplied only by a coefficient that depends on time, not on $x$ itself. Substituting $x = \alpha x_1 + \beta x_2$ and using linearity of differentiation shows every term splits cleanly into an $\alpha$ part and a $\beta$ part, so the whole equation splits into the $u_1$ equation scaled by $\alpha$ plus the $u_2$ equation scaled by $\beta$. Superposition holds, so (i) is linear, its time-dependent coefficient $2t$ does not change this.
Test equation (ii): the term $\left(\frac{dx}{dt}\right)^2$ is the problem. Substituting $x = \alpha x_1 + \beta x_2$ gives
\[ \left(\alpha \dot{x}_1 + \beta \dot{x}_2\right)^2 = \alpha^2 \dot{x}_1^2 + 2\alpha\beta \dot{x}_1\dot{x}_2 + \beta^2 \dot{x}_2^2 \]
which is not the same as $\alpha \dot{x}_1^2 + \beta \dot{x}_2^2$ unless the cross term $2\alpha\beta \dot{x}_1 \dot{x}_2$ happens to vanish, which is not true in general. Superposition fails, so (ii) is not linear.
Test equation (iii): every term is $x$ or a derivative of $x$ with constant coefficients, the same structure as (i) but even simpler since the coefficients do not depend on $t$ either. Superposition holds exactly as in (i), so (iii) is linear.
Step 4: Final Answer.
Applying the superposition test term by term shows only equation (ii) fails, because of its squared derivative term. Equations (i) and (iii) both pass, so the only equation that is not a linear dynamical system is (ii), option (C).
\[ \boxed{\text{(ii) only}} \]
