MCQ
If $u = {e^{ - {x^2} - {y^2}}}$, then
- A$x{u_x} = y{y_y}$
- ✓$y{u_x} = x{u_y}$
- C$y{u_x} + x{u_y} = 0$
- D${x^2}{u_y} + {y^2}{u_x} = 0$
${u_y} = {e^{ - {x^2} - {y^2}}}( - 2y) = - 2uy$
$y{u_x} = x{u_y}$.
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