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Higher Order Derivatives

I f u=e^...

Question

$I$ f $u = \frac{e^{x / y} sin (y / x)}{x^{3}}$
then $x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = ?$

A

0

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B

2 u

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C

12 u

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D

4 u

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Solution

The correct option is C $12 u$
Given, $u = \frac{e^{(\frac{x}{y})} . sin \frac{y}{x}}{x^{3}}$
$u x^{3} = e^{(\frac{x}{y})} . sin \frac{y}{x}$ $\to$ (1)
Partially differentiating both sides w.r.t $x$
$3 x^{2} u + x^{3} \frac{\partial u}{\partial x} = \frac{1}{y} e^{(\frac{x}{y})} sin (\frac{y}{x}) - \frac{y}{x^{2}} e^{(\frac{x}{y})} cos (\frac{y}{x})$
$⟹ 3 x^{2} u + x^{3} \frac{\partial u}{\partial x} = \frac{u x^{3}}{y} - \frac{y}{x^{2}} e^{(\frac{x}{y})} cos (\frac{y}{x})$ $\to$ (2)
Partially differentiating (1) both sides wrt $y$
$x^{3} \frac{\partial u}{\partial y} = \frac{- x}{y^{2}} e^{(\frac{x}{y})} sin (\frac{y}{x}) - \frac{1}{x} e^{(\frac{x}{y})} cos (\frac{y}{x})$
$⟹ x^{3} \frac{\partial u}{\partial y} = \frac{- x^{4} u}{y^{2}} - \frac{1}{x} e^{(\frac{x}{y})} cos (\frac{y}{x})$ $\to$ (3)
(2) $+$ (3) $\times \frac{y}{x}$
$⟹ 3 x^{2} u + x^{3} \frac{\partial u}{\partial x} + x^{2} y \frac{\partial u}{\partial y}$
$⟹ \frac{u x^{3}}{y} - \frac{y}{x^{2}} e^{(\frac{x}{y})} cos (\frac{y}{x}) - \frac{u x^{3}}{y} + \frac{y}{x^{2}} e^{(\frac{x}{y})} cos (\frac{y}{x})$
$∴ 3 x^{2} u + x^{3} \frac{\partial u}{\partial x} + x^{2} y \frac{\partial u}{\partial y} = 0$
$⟹ x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + 3 u = 0$
Partially differentiating both sides wrt $x$ and $y$ separately,
$⟹ x \frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial u}{\partial x} + y \frac{\partial^{2} u}{\partial x \partial y} + 3 \frac{\partial u}{\partial x} = 0$ $\to$ (4)
$⟹ x \frac{\partial^{2} u}{\partial x \partial y} + \frac{\partial u}{\partial y} + y \frac{\partial^{2} u}{\partial^{2} y} + 3 \frac{\partial u}{\partial y} = 0$ $\to$ (5)
$(4)$ $\times x + (5)$ $\times y$
$⟹ x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} + 4 (x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}) = 0$
$∴ ⟹ x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = 12 u$

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Similar questions

u = {tan}^{- 1} (\frac{x^{3} + y^{3}}{x - y})

x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = ?

u = \frac{x}{y^{2}} - \frac{y}{x^{2}}

\frac{\partial^{2} u}{\partial x \partial y} = \frac{\partial^{2} u}{\partial y \partial x}

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\frac{\partial^{2} u}{\partial x^{2}} - \frac{\partial^{2} u}{\partial y^{2}} =

u = f (r)

r^{2} = x^{2} + y^{2} + z^{2}

, then prove that:

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