GATE 2024 EC Q.11: solution and answer

GATE 2024 · ECQ.11 · 1 mark · MCQ

The general form of the complementary function of a differential equation is given by $y (t) = (A t + B) e^{- 2 t}$ , where $A$ and $B$ are real constants determined by the initial condition. The corresponding differential equation is ______.

(A)
$\frac{d ^{2} y}{d t ^{2}} + 4 \frac{d y}{d t} + 4 y = f (t)$
(B)
$\frac{d ^{2} y}{d t ^{2}} + 4 y = f (t)$
(C)
$\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} + 2 y = f (t)$
(D)
$\frac{d ^{2} y}{d t ^{2}} + 5 \frac{d y}{d t} + 6 y = f (t)$

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Answer and solution

ANSWER

The differential equation is $\frac{d ^{2} y}{d t ^{2}} + 4 \frac{d y}{d t} + 4 y = f (t)$ , option (A).

Official answer: ASource: Official answer key, GATE 2024 EC (IISc Bengaluru)

For a linear ODE with constant coefficients, the complementary function (CF) comes from the roots of the auxiliary equation. A CF of the form $(A t + B) e^{m t}$ means $m$ is a repeated (double) root.
Here the CF is $(A t + B) e^{- 2 t}$ , so $m = - 2$ is a root of multiplicity 2.
The auxiliary equation is therefore $(m + 2)^{2} = m^{2} + 4 m + 4 = 0$
Replacing $m$ by the operator $D = \frac{d}{d t}$ gives $\frac{d ^{2} y}{d t ^{2}} + 4 \frac{d y}{d t} + 4 y = f (t),$ which is option (A).
Checking the others: (B) $m^{2} + 4 = 0$ gives $m = \pm 2 j$ , so CF $= A cos 2 t + B sin 2 t$ . (C) $m^{2} + 3 m + 2 = 0$ gives $m = - 1, - 2$ , so CF $= A e^{- t} + B e^{- 2 t}$ . (D) $m^{2} + 5 m + 6 = 0$ gives $m = - 2, - 3$ , so CF $= A e^{- 2 t} + B e^{- 3 t}$ . None of these contains the term $t e^{- 2 t}$ , so only (A) fits.

KEY IDEA

A factor of $t$ multiplying $e^{m t}$ in the complementary function means $m$ is a repeated root of the auxiliary equation. So $(A t + B) e^{- 2 t}$ comes from $(m + 2)^{2} = m^{2} + 4 m + 4 = 0$ . The forcing term $f (t)$ only changes the particular integral, never the CF.

WATCH OUT FOR

Choosing (D) or (C) because each has $- 2$ as a root. They have two distinct roots, so their CF is a sum of two different exponentials with no $t e^{- 2 t}$ term. The other common slip is taking (B) because it shows a 4. But $m^{2} + 4 = 0$ has roots $\pm 2 j$ , which give an oscillating CF, not a decaying one.

From the GATE 2024 EC paper. Every question of this paper · Download the full paper · All 19 Differential equations questions, with notes

GATE 2024 EC Q.11: solution and answer

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