Q98 / 200 · AP EAMCET 2001 · Mathematics

If $u=e^{x^2-y^2}$, then

  • A
    $x u_x=y u_y$
  • B
    $y u_x=x u u_y$
  • Correct answer
    C
    $y u_x+x u_y=0$
  • D
    $x^2 u_y+y^2 u_x=0$

Solution

Given that, $u=e^{x^2-y^2}$ On differentiating partially $u_x=e^{x^2-y^2}(2 x)$ On differentiating partially w.r.t. $y$. $\begin{aligned} u_y & =e^{x^2-y^2}(-2 y) \\ y u_x & =e^{x^2-y^2} 2 x y \\ x u_y & =e^{x^2-y^x}(-2 x y)\end{aligned}$ On adding Eqs. (i) and (ii) $y u_x+x u_y=0$

Hence, the correct answer is option: 'C'.

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