If $u \equiv u(x, y)=\sin (y+a x)-(y+a x)^2$, then it implies
- $u_{x x}=a^2 \cdot u_{y y}$
- $u_{y y}=a^2 u_{x x}$
- $u_{x x}=-a^2 \cdot u_{y y}$
- $u_{y y}=-a^2 u_{x x}$
Solution

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 $u_{x x}=a^2 u_{y y}$
Asked in: AP EAMCET 2011