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If $u=e^{x^2-y^2}$, then
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Verified Answer
The correct answer is:
$y u_x+x u_y=0$
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)
$y u_x+x u_y=0$
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)
$y u_x+x u_y=0$
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