If u equiv u(x, y) = sin(y + ax) - (y + ax)^2 | vedclass

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(A) Given $u = \sin(y + ax) - (y + ax)^2$ ... $(i)$Differentiating partially with respect to $x$:$u_x = \cos(y + ax) \cdot a - 2(y + ax) \cdot a$$u_x

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$u_{xx} = a^2 \cdot u_{yy}$

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(A) Given $u = \sin(y + ax) - (y + ax)^2$ ... $(i)$Differentiating partially with respect to $x$:$u_x = \cos(y + ax) \cdot a - 2(y + ax) \cdot a$$u_x = a[\cos(y + ax) - 2(y + ax)]$Differentiating again with respect to $x$:$u_{xx} = a[-\sin(y + ax) \cdot a - 2 \cdot a]$$u_{xx} = -a^2[\sin(y + ax) + 2]$ ... (ii)Differentiating $(i)$ partially with respect to $y$:$u_y = \cos(y + ax) - 2(y + ax)$Differentiating again with respect to $y$:$u_{yy} = -\sin(y + ax) - 2$$u_{yy} = -[\sin(y + ax) + 2]$ ... (iii)From (ii) and (iii),we can see that:$u_{xx} = a^2 \cdot [-\sin(y + ax) - 2]$$u_{xx} = a^2 \cdot u_{yy}$

If $u \equiv u(x, y) = \sin(y + ax) - (y + ax)^2$,then it implies

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