Consider z=(y+it)+(y-it), where z,y,t are variables and i=-1. The partial differential equation derived from this expression is

GATE 2022 CE (CE1) – Question 11

Engineering Mathematics · PDE: Fourier series, separation of variables, diffusion, wave and Laplace equations · 1 mark · Multiple choice

Consider $z=\sin(y+it)+\cos(y-it)$, where z,y,t are variables and $i=\sqrt{-1}$. The partial differential equation derived from this expression is

$z_{tt}+z_{yy}=0$
$z_{tt}-z_{yy}=0$
$z_t-iz_y=0$
$z_t+iz_y=0$

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Correct answer: (A) $z_{tt}+z_{yy}=0$

Explanation

Each term is a function of $y\pm it$. Differentiating twice in t contributes $i^2=-1$ relative to twice differentiating in y.

Thus $z_{tt}=-z_{yy}$ and $\mathbf{z_{tt}+z_{yy}=0}$, **A**. The first derivatives do not obey a single sign relation because the two terms use opposite signs of it.