Question

Re- correction

$\bf \: if \: \rm \: y = { (x + \sqrt{1 + {x}^{2} } } )^{ \: m} \\ \\ \bf \: then \: prove \: that : \\ \\ \rm \: {(1 + {x}^{2} ). \frac{ {d}^{2}y}{d {x}^{2} } }^{} + x \frac{dy}{dx} - {m}^{2} y = 0 \\ \\ \\ \bf \: and \: also \: find \: the \: value \: of \: \: \frac{ {d}^{3}y }{d {x}^{3} } \\ \\ \bf \: when \: \: x = 0$
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Answer 1

This is your answer.

Hope it will help you..

Answer 2

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

Given that

$\rm :\longmapsto\:y = {(x + \sqrt{ {x}^{2} + 1}) }^{m}$

On taking log both sides, we get

$\rm :\longmapsto\:logy = log {(x + \sqrt{ {x}^{2} + 1}) }^{m}$

We know,

$\rm :\longmapsto\:\boxed{ \tt{ \: \: log {x}^{y} = ylogx \: \: }}$

So, using this, we get

$\rm :\longmapsto\:logy \: = \: m \: log[x + \sqrt{ {x}^{2} + 1 } ]$

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

$\rm :\longmapsto\:\dfrac{d}{dx}logy \: = \: m \:\dfrac{d}{dx} log[x + \sqrt{ {x}^{2} + 1 } ] \:$

We know,

$\rm :\longmapsto\:\boxed{ \tt{ \: \: \dfrac{d}{dx}logx = \frac{1}{x} \: \: }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} =m \times \dfrac{1 }{x + \sqrt{ {x}^{2} + 1 }} \dfrac{d}{dx}[x + \sqrt{ {x}^{2} + 1 }]$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} =\dfrac{m}{x + \sqrt{ {x}^{2} + 1 }} \dfrac{d}{dx}[x + \sqrt{ {x}^{2} + 1 }]$

We know,

$\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{d}{dx}x = 1 \: }} \: \: and \: \: \boxed{ \tt{ \: \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} } \: }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} =\dfrac{m}{x + \sqrt{ {x}^{2} + 1 }}\bigg[1 + \dfrac{1}{2 \sqrt{ {x}^{2} + 1}} \dfrac{d}{dx}( {x}^{2} + 1) \bigg]$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} =\dfrac{m}{x + \sqrt{ {x}^{2} + 1 }}\bigg[1 + \dfrac{1}{2 \sqrt{ {x}^{2} + 1}} \times 2x \bigg]$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} =\dfrac{m}{x + \sqrt{ {x}^{2} + 1 }}\bigg[1 + \dfrac{x}{\sqrt{ {x}^{2} + 1}} \bigg]$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} =\dfrac{m}{x + \sqrt{ {x}^{2} + 1 }}\bigg[\dfrac{ \sqrt{ {x}^{2} + 1 } + x}{\sqrt{ {x}^{2} + 1}} \bigg]$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} =\dfrac{m}{\sqrt{ {x}^{2} + 1 }}$

$\rm :\longmapsto\: \sqrt{ {x}^{2} + 1} \: \dfrac{dy}{dx} = my$

On squaring both sides, we get

$\rm :\longmapsto\:( {x}^{2} + 1) {\bigg[\dfrac{dy}{dx} \bigg]}^{2} = {m}^{2} {y}^{2}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}( {x}^{2} + 1) {\bigg[\dfrac{dy}{dx} \bigg]}^{2} = {m}^{2} \dfrac{d}{dx}{y}^{2}$

$\rm :\longmapsto\: ({x}^{2} + 1)\dfrac{d}{dx} {\bigg[\dfrac{dy}{dx} \bigg]}^{2} + {\bigg[\dfrac{dy}{dx} \bigg]}^{2}\dfrac{d}{dx}( {x}^{2} + 1) = 2y{m}^{2} \dfrac{dy}{dx}$

$\rm :\longmapsto\: 2({x}^{2} + 1)\dfrac{dy}{dx} \dfrac{ {d}^{2}y}{d {x}^{2} } + {\bigg[\dfrac{dy}{dx} \bigg]}^{2}(2x) = 2y{m}^{2} \dfrac{dy}{dx}$

$\rm :\longmapsto\: 2\dfrac{dy}{dx} \bigg[( {x}^{2} + 1) \dfrac{ {d}^{2}y}{d {x}^{2} } + x{\bigg[\dfrac{dy}{dx} \bigg]}\bigg] = 2y{m}^{2} \dfrac{dy}{dx}$

$\rm :\longmapsto\: ( {x}^{2} + 1) \dfrac{ {d}^{2}y}{d {x}^{2} } + x{\bigg[\dfrac{dy}{dx} \bigg]} = y{m}^{2}$

$\rm :\longmapsto\: ( {x}^{2} + 1) \dfrac{ {d}^{2}y}{d {x}^{2} } + x{\bigg[\dfrac{dy}{dx} \bigg]} = {m}^{2} y$

$\rm :\longmapsto\: ( {x}^{2} + 1) \dfrac{ {d}^{2}y}{d {x}^{2} } + x{\bigg[\dfrac{dy}{dx} \bigg]} - {m}^{2} y = 0$

Hence, Proved

When x = 0

$\rm :\longmapsto\:y = 1$

$\rm :\longmapsto\:\dfrac{dy}{dx} = m$

$\rm :\longmapsto\:\dfrac{ {d}^{2} y}{d {x}^{2} } = {m}^{2}$

Now, we have

$\rm :\longmapsto\: ( {x}^{2} + 1) \dfrac{ {d}^{2}y}{d {x}^{2} } + x{\bigg[\dfrac{dy}{dx} \bigg]} - {m}^{2} y = 0$

can be rewritten as

$\rm :\longmapsto\:( {x}^{2} + 1)y_2 + xy_1 - {m}^{2} y = 0$

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

$\rm :\longmapsto\:\dfrac{d}{dx}( {x}^{2} + 1)y_2 + \dfrac{d}{dx}xy_1 - \dfrac{d}{dx} {m}^{2} y = 0$

$\rm :\longmapsto\:( {x}^{2} + 1)y_3 + y_2(2x) + xy_2 + y_1 - {m}^{2}y_1 = 0$

Now, Substitute x = 0, we get

$\rm :\longmapsto\:(0 + 1)y_3 + m - {m}^{2} \times m = 0$

$\rm :\longmapsto\:y_3 + m - {m}^{3} = 0$

$\bf\implies \:y = {m}^{3} - m$

$\bf\implies \:\dfrac{ {d}^{3} y}{ {dx}^{3} } = m( {m}^{2} - 1)$