Find a possible solution of / x=-2/ t and / x=-2/ t.

$\alpha=(x-t)^2+(x+t)^2,\ \beta=(x-t)^2-(x+t)^2$

$\alpha=(x-2t)^2+(x+2t)^2,\ \beta=(x-2t)^2-(x+2t)^2$

$\alpha=(x-t/2)^2+2(x+t/2)^2,\ \beta=2(x-t/2)^2-(x+t/2)^2$

GATE 2024 BM – Question 37

Engineering Mathematics · Differential Equations · 2 marks · Multiple choice

Find a possible solution of $\partial\alpha/\partial x=-2\partial\beta/\partial t$ and $\partial\beta/\partial x=-2\partial\alpha/\partial t$.

$\alpha=(x-t)^2+(x+t)^2,\ \beta=(x-t)^2-(x+t)^2$
$\alpha=(x-2t)^2+(x+2t)^2,\ \beta=(x-2t)^2-(x+2t)^2$
$\alpha=(x-t/2)^2+(x+t/2)^2,\ \beta=(x-t/2)^2-(x+t/2)^2$
$\alpha=(x-t/2)^2+2(x+t/2)^2,\ \beta=2(x-t/2)^2-(x+t/2)^2$

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Correct answer: (C) $\alpha=(x-t/2)^2+(x+t/2)^2,\ \beta=(x-t/2)^2-(x+t/2)^2$

Explanation

For C, expanding gives α=2x²+t²/2 and β=−2xt. Then αx=4x and −2βt=4x; also βx=−2t and −2αt=−2t. Both equations hold identically. Thus C supplies the required solution.

GATE 2024 BM – Question 37

Explanation

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