Question

Prove it........ #### its urgent ​

Answer 1

Given---->

$y \: = {e}^{m \: {sin}^{ - 1}x }$

To prove ---->

$( \: 1 - {x}^{2} ) \: \dfrac{ {d}^{2}y }{d {x}^{2} } \: - \: x \dfrac{dy}{dx} \: = {m}^{2} y$

Concept used ---->

1)

$\dfrac{d}{dx} ( \: {e}^{x} \: ) = {e}^{x}$

2)

$\dfrac{d}{dx} \: ( \: cos ^{ - 1} x \: ) \: = - \dfrac{1}{ \sqrt{1 - {x}^{2} } }$

3)

$\dfrac{d}{dx} \: ( \: u \: v \: ) \: = u \: \dfrac{dv}{dx} \: + \: v \: \dfrac{du}{dx}$

4)

$\dfrac{d}{dx} ( {x}^{2} ) = 2x$

5)

$y \: ( \: 1 \: ) \: = \dfrac{dy}{dx}$

$y \: ( \: 2 \: ) = \dfrac{d^{2}y }{d {x}^{2} }$

Solution---->

$y \: = {e}^{m \: {cos}^{ - 1} x}$

$differentiting \: with \: respect \: to \: x \: we \: get$

$\dfrac{dy}{dx} \: = \dfrac{d}{dx} \: ( \: {e}^{m {cos}^{ - 1}x } \: )$

$\dfrac{dy}{dx} \: = {e}^{m {cos}^{ - 1} x} \dfrac{d}{dx} \: (m \: {cos}^{ - 1} x \: )$

$\dfrac{dy}{dx} \: = {e}^{m {cos}^{ - 1}x } \: ( \: - \dfrac{m}{ \sqrt{1 - {x}^{2} } } \: )$

$= > \sqrt{1 - {x}^{2} } \: \: \dfrac{dy}{dx} \: = - m \: {e}^{m \: {cos}^{ - 1} x}$

$squaring \: both \: sides \: we \: get \:$

$= > ( \: 1 \: - {x}^{2} \: ) \: ( \dfrac{dy}{dx} \: )^{2} \: = {m}^{2} \: ( \: {e}^{m {cos}^{ - 1} x} \: ) ^{2}$

$putting \: y \: = \: {e}^{m {cos}^{ - 1}x } \: we \: get$

$= > ( \: 1 - {x}^{2} \: ) \: ({ \frac{dy}{dx}) }^{2} = {m}^{2} \: {y}^{2}$

$differentiting \: with \: respect \: to \: x \: we \: get \:$

$= > \dfrac{d}{dx} ( \: \: (1 - {x}^{2} ) \: { (\dfrac{dy}{dx}) }^{2} ) = {m}^{2} \dfrac{d}{dx} ( {y}^{2} )$

$= > (1 - {x}^{2} ) \: \dfrac{d}{dx} ( { \dfrac{dy}{dx} )}^{2} \: + { (\dfrac{dy}{dx} )}^{2} \: \dfrac{d}{dx} \: (1 - {x}^{2} ) = 2 {m}^{2} y \: \dfrac{dy}{dx}$

$= > (1 - {x}^{2} ) \: 2 \: \dfrac{dy}{dx} \: \: \dfrac{ {d}^{2} y}{d {x}^{2} } + ({ \dfrac{dy}{dx}) }^{2} \: (0 - 2x) = 2 {m}^{2} y \: \dfrac{dy}{dx}$

$= > (1 - {x}^{2} ) \: 2 \: \dfrac{dy}{dx} \: \dfrac{ {d}^{2}y }{d {x}^{2} } - 2x( \dfrac{dy}{dx} ) ^{2} = 2 {m}^{2} y \: \dfrac{dy}{dx}$

$\: divided \: whole \: equation \: by \: 2 \: \dfrac{dy}{dx} \\ we \: get$

$= > (1 - {x}^{2} ) \dfrac{ {d}^{2}y }{d {x}^{2} } \: - x \: \dfrac{dy}{dx} \: = {m}^{2} y$

$= > (1 - {x}^{2} ) \: y(2) \: - x \: y(1) = \: {m}^{2} y$

Answer 2

$\huge\boxed{\fcolorbox{pink}{blue}{Answer}}$