Question

$\rm If \: z = {x}^2 + y^2 ,prove \: that \\ \rm{x}^{2} \frac{ \partial^{2} z}{ \partial { x}^{2} } +2 x y\frac{ \partial {}^{2} z}{\partial x\partial y} + {y}^{2} \frac{ {\partial}^{2}z }{\partial {y}^{2} } = 2z$​

Answer 1

Solution:

Given That:

$\rm \longrightarrow z = {x}^{2} + {y}^{2}$

Partial differentiating both sides with respect to x, we get:

$\rm \longrightarrow \dfrac{ \partial z}{ \partial x} = 2x$

Partial differentiating again with respect to x, we get:

$\rm \longrightarrow \dfrac{ \partial^{2} z}{ \partial x^{2} } = 2 -(i)$

Consider again:

$\rm \longrightarrow z = {x}^{2} + {y}^{2}$

Partial differentiating both sides with respect to y, we get:

$\rm \longrightarrow \dfrac{ \partial z}{ \partial y} = 2y$

Now, partial differentiating both sides with respect to x, we get:

$\rm \longrightarrow \dfrac{ \partial^{2} z}{ \partial x \: \partial y} = 0 - (ii)$

Consider again:

$\rm \longrightarrow \dfrac{ \partial z}{ \partial y} = 2y$

Partial differentiating both sides with respect to y, we get:

$\rm \longrightarrow \dfrac{ \partial^{2} z}{ \partial y^{2} } = 2 - (iii)$

Now, consider Left Hand Side, we get:

$\rm = {x}^{2} \dfrac{ \partial^{2} z}{ \partial {x}^{2} } + 2xy \dfrac{ { \partial}^{2}x}{ \partial x \: \partial y} + {y}^{2} \dfrac{ \partial^{2}z }{ \partial {y}^{2} }$

From (i), (ii) and (iii), we can write:

$\rm = 2{x}^{2} + 0 + 2{y}^{2}$

$\rm = 2({x}^{2} + {y}^{2} )$

$\rm = 2z$

Therefore:

$\rm \longrightarrow {x}^{2} \dfrac{ \partial^{2} z}{ \partial {x}^{2} } + 2xy \dfrac{ { \partial}^{2}x}{ \partial x \: \partial y} + {y}^{2} \dfrac{ \partial^{2}z }{ \partial {y}^{2}} = 2z$

Hence Proved.

Learn More:

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