GATE 2018 IN Q.29: solution and answer

GATE 2018 · INQ.29 · 2 marks · MCQ

Consider the following equations

\frac{\partial V ( x , y )}{\partial x} = p x^{2} + y^{2} + 2 x y

\frac{\partial V ( x , y )}{\partial y} = x^{2} + q y^{2} + 2 x y

where $p$ and $q$ are constants. $V (x, y)$ that satisfies the above equations is

(A)
$p \frac{x ^{3}}{3} + q \frac{y ^{3}}{3} + 2 x y + 6$
(B)
$p \frac{x ^{3}}{3} + q \frac{y ^{3}}{3} + 5$
(C)
$p \frac{x ^{3}}{3} + q \frac{y ^{3}}{3} + x^{2} y + x y^{2} + x y$
(D)
$p \frac{x ^{3}}{3} + q \frac{y ^{3}}{3} + x^{2} y + x y^{2}$

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Answer and solution

ANSWER

$V (x, y) = p \frac{x ^{3}}{3} + q \frac{y ^{3}}{3} + x^{2} y + x y^{2}$ (up to an additive constant), which is option (D).

Official answer: DSource: Official answer key, GATE 2018 IN (IIT Guwahati)

Consistency check (equality of mixed partials). A $V$ can exist only if $\frac{\partial}{\partial y} (\frac{\partial V}{\partial x}) = \frac{\partial}{\partial x} (\frac{\partial V}{\partial y})$ . Here $\frac{\partial}{\partial y} (p x^{2} + y^{2} + 2 x y) = 2 y + 2 x$ and $\frac{\partial}{\partial x} (x^{2} + q y^{2} + 2 x y) = 2 x + 2 y$ . They agree for every $p$ and $q$ , and both right-hand sides are polynomials defined on the whole plane, so such a $V$ exists.
Integrate the first equation in $x$ , holding $y$ fixed: $V = p \frac{x ^{3}}{3} + x y^{2} + x^{2} y + g (y)$ where $g (y)$ is the 'constant' of integration, which may depend on $y$ .
Differentiate in $y$ and match the second equation: $\frac{\partial V}{\partial y} = 2 x y + x^{2} + g^{'} (y) = x^{2} + q y^{2} + 2 x y \Rightarrow g^{'} (y) = q y^{2}$
Integrate: $g (y) = q \frac{y ^{3}}{3} + C$ , so $V (x, y) = p \frac{x ^{3}}{3} + q \frac{y ^{3}}{3} + x^{2} y + x y^{2} + C$ where $C$ is an arbitrary constant.
Match the options. With $C = 0$ this is option (D). Check: for (D), $\partial V / \partial x = p x^{2} + 2 x y + y^{2}$ and $\partial V / \partial y = q y^{2} + x^{2} + 2 x y$ , both as given. The others fail the first equation: (A) gives $\partial V / \partial x = p x^{2} + 2 y$ , (B) gives $p x^{2}$ , and (C) gives $p x^{2} + 2 x y + y^{2} + y$ , where the extra $x y$ term adds a stray $y$ . The constants $6$ and $5$ in (A) and (B) are not the problem, since a constant never changes a partial derivative. Those options fail because they are missing the $x^{2} y + x y^{2}$ terms.

KEY IDEA

To find $V$ from its two partial derivatives, integrate one of them, letting the 'constant' of integration be a function of the other variable. Then differentiate that result with respect to the other variable and match it to the second given partial to fix that function. For an MCQ, the quickest check is to differentiate each option and compare with both given partials.

WATCH OUT FOR

One mistake is to integrate each equation on its own and add the results. That counts the shared cross terms twice, giving $2 x^{2} y + 2 x y^{2}$ instead of $x^{2} y + x y^{2}$ . Another is to treat the $2 x y$ term as if it came from an $x y$ (or $2 x y$ ) term in $V$ , which leads to (A) or (C). The right source is $x^{2} y + x y^{2}$ : differentiating it in $x$ gives $2 x y + y^{2}$ , and in $y$ gives $x^{2} + 2 x y$ . A third mistake is to use a true constant instead of $g (y)$ as the constant of integration after integrating in $x$ .

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GATE 2018 IN Q.29: solution and answer

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