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If $u \equiv u(x, y)=\sin (y+a x)-(y+a x)^2$, then it implies
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Verified Answer
The correct answer is:
$u_{x x}=a^2 \cdot u_{y y}$
On differentiating partially w.r.t. $x$, we get
$u_x=\cos (y+a x) a-2(y+a x) a$
Again differentiating partially w.r.t. $x$, we get
On differentiating partially Eq. (i) w.r.t. $y$, we get
$u_y=\cos (y+a x)-2(y+a x)$
$\therefore$ From Eqs. (ii) and (iii), we get
$u_{x x}=a^2 u_{y y}$
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