Question

y=e^asin-1x show that (1-x^2)y2-xy1-a^2y=0​

Answer 1

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

Given function is

$\rm :\longmapsto\:y = {e}^{a \: {sin}^{ - 1} x}$

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

$\rm :\longmapsto\:\dfrac{d}{dx} y =\dfrac{d}{dx} {e}^{a \: {sin}^{ - 1} x}$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx} {e}^{x} = {e}^{x} \: }}$

So, using this

$\rm :\longmapsto\:y_1 = {e}^{a \: {sin}^{ - 1} x} \dfrac{d}{dx} {a \: sin}^{ - 1}x$

$\rm :\longmapsto\:y_1 =a \: y \dfrac{d}{dx} { \: sin}^{ - 1}x$

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

$\rm :\longmapsto\: \sqrt{1 - {x}^{2}} \: y_1 \: = \: ay$

On squaring both sides, we get

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

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

$\rm :\longmapsto\: \dfrac{d}{dx}{(1 - x}^{2}) {y_1}^{2} = \dfrac{d}{dx}{a}^{2} {y}^{2}$

We know,

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

So, using this, we get

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

$\rm :\longmapsto\:(1 - {x}^{2})2y_1y_2 + {y_1}^{2}(0 - 2x) = {a}^{2}2yy_1$

$\rm :\longmapsto\:(1 - {x}^{2})2y_1y_2 - 2x {y_1}^{2}= {a}^{2}2yy_1$

$\rm :\longmapsto\:2y_1 \: [(1 - {x}^{2})y_2 - x y_1]= 2{a}^{2}yy_1$

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

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

Hence, Proved

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More to know :-

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