Question

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$x = tan( \frac{1}{a} logy)$

Prove that,

$( {x}^{2} + {a}^{2} )y_2 + (2x - a)y_1 = 0$
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Answer 1

Appropriate Question :-

If

$\rm :\longmapsto\:x = tan\bigg[\dfrac{1}{a}logy\bigg]$

Prove that,

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

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

Given that,

$\rm :\longmapsto\:x = tan\bigg[\dfrac{1}{a}logy\bigg]$

can be rewritten as

$\rm :\longmapsto\: {tan}^{ - 1}x = \dfrac{1}{a}logy$

$\rm :\longmapsto\: a{tan}^{ - 1}x = logy$

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

$\rm :\longmapsto\: \dfrac{d}{dx} a{tan}^{ - 1}x =\dfrac{d}{dx} logy$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx} {tan}^{ - 1}x = \frac{1}{ {x}^{2} + 1}}}$

and

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

So, on substituting the values, we get

$\rm :\longmapsto\:a \times \dfrac{1}{ {x}^{2} + 1} = \dfrac{1}{y}\dfrac{d}{dx}y$

$\rm :\longmapsto\: \dfrac{a}{ {x}^{2} + 1} = \dfrac{1}{y}y_1$

$\bf\implies \:\boxed{ \tt{ \: ( {x}^{2} + 1)y_1 = ay \: }}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}( {x}^{2} + 1)y_1 = \dfrac{d}{dx}ay$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}u.v \: = \: u\dfrac{d}{dx}v +v\dfrac{d}{dx}u \: }}$

So, using this, we get

$\rm :\longmapsto\:( {x}^{2} + 1)\dfrac{d}{dx}y_1 + y_1\dfrac{d}{dx}( {x}^{2} + 1) = ay_1$

We know,

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

So, using this, we get

$\rm :\longmapsto\:( {x}^{2} + 1)y_2 + 2xy_1 = ay_1$

$\rm :\longmapsto\:( {x}^{2} + 1)y_2 + 2xy_1 - ay_1 = 0$

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: ( {x}^{2} + 1)y_2 + (2x - a)y_1 = 0}}}$

Hence, Proved

More to know :-

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