GATE EC 2024 Set 1 — Question 11

The complementary function of a linear homogeneous differential equation with constant coefficients is given by $y_c(t)$.
Given $y_c(t) = (At + B)e^{-2t}$.This form of the complementary function indicates that the characteristic equation of the differential equation has a repeated root at $s = -2$.If $s = -2$ is a repeated root, the characteristic equation must be of the form $(s - (-2))^2 = 0$.
$(s + 2)^2 = 0$
$s^2 + 4s + 4 = 0$The corresponding homogeneous differential equation is obtained by replacing $s^2$ with $\frac{d^2}{dt^2}$, $s$ with $\frac{d}{dt}$, and the constant term with the coefficient of $y(t)$.
So, the homogeneous part of the differential equation is $\frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 4y = 0$.For a non-homogeneous differential equation, the right-hand side is a function of $t$, typically denoted as $f(t)$.
Thus, the corresponding differential equation is $\frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 4y = f(t)$.Comparing this with the given options, option (A) matches this form.Options (B), (C), and (D) would have different characteristic equations and thus different complementary functions:
(B) $s^2 + 4 = 0 \implies s = \pm 2j \implies y_c(t) = C_1 \cos(2t) + C_2 \sin(2t)$
(C) $s^2 + 3s + 2 = 0 \implies (s+1)(s+2) = 0 \implies s = -1, -2 \implies y_c(t) = C_1 e^{-t} + C_2 e^{-2t}$
(D) $s^2 + 5s + 6 = 0 \implies (s+2)(s+3) = 0 \implies s = -2, -3 \implies y_c(t) = C_1 e^{-2t} + C_2 e^{-3t}$Therefore, option (A) is the correct differential equation.The final answer is $\boxed{\text{A}}$