GATE 2022 EC Q.25: solution and answer

GATE 2022 · ECQ.25 · 1 mark · MSQ

Consider the following partial differential equation (PDE)

a \frac{\partial ^{2} f ( x , y )}{\partial x ^{2}} + b \frac{\partial ^{2} f ( x , y )}{\partial y ^{2}} = f (x, y),

where $a$ and $b$ are distinct positive real numbers. Select the combination(s) of values of the real parameters $ξ$ and $η$ such that $f (x, y) = e^{(ξ x +)}$ is a solution of the given PDE.

[As printed, the second term of the exponent is blank in the paper. Since the options give values for $η$ , the intended function is $f (x, y) = e^{(ξ x + η y)}$ .]

(A)
$ξ = \frac{1}{2 a}, η = \frac{1}{2 b}$
(B)
$ξ = \frac{1}{a}, η = 0$
(C)
$ξ = 0, η = 0$
(D)
$ξ = \frac{1}{a}, η = \frac{1}{b}$

See the question in the paper

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Answer and solution

ANSWER

The correct options are (A) and (B). Substituting $f = e^{ξ x + η y}$ gives the condition $a ξ^{2} + b η^{2} = 1$ , and only options (A) and (B) satisfy it.

Official answer: A, BSource: Official answer key, GATE 2022 EC (IIT Kharagpur)

Try the exponential solution $f = e^{ξ x + η y}$ . The blank in the printed exponent has to be $η y$ , because the options give values of $η$ . Each derivative brings down a constant factor: $\frac{\partial f}{\partial x} = ξ f$ , $\frac{\partial ^{2} f}{\partial x ^{2}} = ξ^{2} f$ and $\frac{\partial ^{2} f}{\partial y ^{2}} = η^{2} f$ .
Substitute these into the PDE: $a ξ^{2} f + b η^{2} f = f$ . Since $e^{ξ x + η y} \neq = 0$ for every $x$ and $y$ , we can divide by $f$ . This gives the condition $a ξ^{2} + b η^{2} = 1.$
(A): $a \cdot \frac{1}{2 a} + b \cdot \frac{1}{2 b} = \frac{1}{2} + \frac{1}{2} = 1$ . The condition holds, so (A) is correct.
(B): $a \cdot \frac{1}{a} + b \cdot 0 = 1$ . The condition holds, so (B) is correct.
(C): $a \cdot 0 + b \cdot 0 = 0 \neq = 1$ , so it fails. Here $f = e^{0} = 1$ , which makes the left side $0$ while the right side is $1$ .
(D): $a \cdot \frac{1}{a} + b \cdot \frac{1}{b} = 1 + 1 = 2 \neq = 1$ , so it fails.
The correct options are therefore (A) and (B). In general, any pair $(ξ, η)$ on the ellipse $a ξ^{2} + b η^{2} = 1$ gives a solution. It is an ellipse because $a, b > 0$ .

KEY IDEA

With an exponential trial solution $e^{ξ x + η y}$ , each second partial derivative becomes a multiplication: by $ξ^{2}$ for $x$ and by $η^{2}$ for $y$ . The PDE then reduces to one algebraic condition, $a ξ^{2} + b η^{2} = 1$ , and each option is checked against it.

WATCH OUT FOR

One mistake is to pick (D) by requiring each term to match $f$ on its own ( $a ξ^{2} = 1$ and $b η^{2} = 1$ ). Only the sum has to equal $1$ , and (D) makes it $2$ . Another is to pick (C) because a zero exponent looks like the "trivial" solution. But $e^{0} = 1$ , not the zero function, and the constant $1$ does not satisfy $0 = 1$ .

From the GATE 2022 EC paper. Every question of this paper · Download the full paper · All 19 Differential equations questions, with notes

GATE 2022 EC Q.25: solution and answer

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