Question

$\rm If \: y = {x}^3 + \frac{1}{ {x}^{3}} ,prove \: that \\ \rm{x}^{2} \frac{ {d}^{2} y}{d {x }^{2} } + x \frac{dy}{dx} = 9y$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: y = {x}^3 + \dfrac{1}{ {x}^{3}}$

can be rewritten as

$\rm \: y = {x}^{3} + {x}^{ - 3}$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx}y = \dfrac{d}{dx}( {x}^{3} + {x}^{ - 3})$

$\rm \: \dfrac{dy}{dx} = {3x}^{3 - 1} - {3x}^{ - 3 - 1}$

$\red{\rm\implies \:\dfrac{dy}{dx} = {3x}^{2} - {3x}^{ - 4}} - - - (1)$

Now, Again on differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx}\bigg(\dfrac{dy}{dx}\bigg) = \dfrac{d}{dx}( {3x}^{2} - {3x}^{ - 4})$

$\rm \: \dfrac{d^{2}y }{d {x}^{2} } = {6x}^{2 - 1} - 3( - 4) {x}^{ - 4 - 1}$

$\red{\rm\implies \:\rm \: \dfrac{d^{2}y }{d {x}^{2} } =6x + 12 {x}^{ - 5}} - - - (2)$

Now, Consider

$\: \rm \: \: {x}^{2} \dfrac{ {d}^{2} y}{d {x }^{2} } + x \dfrac{dy}{dx}$

On substituting the values from equation (1) and (2), we get

$\rm \: = \: {x}^{2}(6x + 12 {x}^{ - 5}) + x( {3x}^{2} - {3x}^{ - 4})$

$\rm \: = \: {6x}^{3} + 12 {x}^{ - 3} + {3x}^{3} - {3x}^{ - 3}$

$\rm \: = \: {9x}^{3} +9{x}^{ - 3}$

$\rm \: = \: 9( {x}^{3} + {x}^{ - 3})$

$\rm \: = \:9 \bigg({x}^3 + \dfrac{1}{ {x}^{3}}\bigg)$

$\rm \: = \: 9y$

Hence,

$\rm\implies \:\boxed{ \bf{ \: {x}^{2} \dfrac{ {d}^{2} y}{d {x }^{2} } + x \dfrac{dy}{dx} = 9y \: \: }}$

$\rule{190pt}{2pt}$

Formulae Used :-

$\boxed{ \rm{ \:\dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} \: \: }}$

$\boxed{ \rm{ \: {a}^{m} \times {a}^{n} = {a}^{m + n} \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$