Question

y = sin (msin^-1x)
prove
(1-x^2)d2y/dx^2 - xdy/dc + m^2y = 0​

Answer 1

Note that $y_1=\dfrac{dy}{dx}$ and $y_2=\dfrac{d^2y}{dx^2}.$

Given,

$\longrightarrow y=\sin(m\sin^{-1}x)$

$\longrightarrow m^2y=m^2\sin(m\sin^{-1}x)\quad\quad\dots(1)$

First derivative of $y$ is,

$\longrightarrow y_1=\dfrac{d}{dx}\left[\sin(m\sin^{-1}x)\right]$

$\longrightarrow y_1=\cos(m\sin^{-1}x)\cdot \dfrac{m}{\sqrt{1-x^2}}$

$\longrightarrow y_1=\dfrac{m\cos(m\sin^{-1}x)}{\sqrt{1-x^2}}$

$\longrightarrow xy_1=\dfrac{mx\cos(m\sin^{-1}x)}{\sqrt{1-x^2}}\quad\quad\dots(2)$

Second derivative of $y$ is,

$\longrightarrow y_2=\dfrac{d}{dx}\left[\dfrac{m\cos(m\sin^{-1}x)}{\sqrt{1-x^2}}\right]$

$\longrightarrow y_2=m\cdot\dfrac{-\sin(m\sin^{-1}x)\cdot\dfrac{m}{\sqrt{1-x^2}}\cdot\sqrt{1-x^2}-\cos(m\sin^{-1}x)\cdot\dfrac{-2x}{2\sqrt{1-x^2}}}{1-x^2}$

$\longrightarrow y_2=\dfrac{mx\cos(m\sin^{-1}x)-m^2\sqrt{1-x^2}\sin(m\sin^{-1}x)}{(1-x^2)\sqrt{1-x^2}}$

$\longrightarrow(1-x^2)y_2=\dfrac{mx\cos(m\sin^{-1}x)-m^2\sqrt{1-x^2}\sin(m\sin^{-1}x)}{\sqrt{1-x^2}}$

$\longrightarrow(1-x^2)y_2=\dfrac{mx\cos(m\sin^{-1}x)}{\sqrt{1-x^2}}-m^2\sin(m\sin^{-1}x)$

From (1) and (2),

$\longrightarrow(1-x^2)y_2=xy_1-m^2y$

$\longrightarrow\underline{\underline{(1-x^2)y_2-xy_1+m^2y=0}}$

Hence Proved!

Answer 2

Answer:

Note that $y_1=\dfrac{dy}{dx}$ and $y_2=\dfrac{d^2y}{dx^2}.$

Given,

$\longrightarrow y=\sin(m\sin^{-1}x)$

$\longrightarrow m^2y=m^2\sin(m\sin^{-1}x)\quad\quad\dots(1)$

First derivative of $y$ is,

$\longrightarrow y_1=\dfrac{d}{dx}\left[\sin(m\sin^{-1}x)\right]$

$\longrightarrow y_1=\cos(m\sin^{-1}x)\cdot \dfrac{m}{\sqrt{1-x^2}}$

$\longrightarrow y_1=\dfrac{m\cos(m\sin^{-1}x)}{\sqrt{1-x^2}}$

$\longrightarrow xy_1=\dfrac{mx\cos(m\sin^{-1}x)}{\sqrt{1-x^2}}\quad\quad\dots(2)$

Second derivative of $y$ is,

$\longrightarrow y_2=\dfrac{d}{dx}\left[\dfrac{m\cos(m\sin^{-1}x)}{\sqrt{1-x^2}}\right]$

$\longrightarrow y_2=m\cdot\dfrac{-\sin(m\sin^{-1}x)\cdot\dfrac{m}{\sqrt{1-x^2}}\cdot\sqrt{1-x^2}-\cos(m\sin^{-1}x)\cdot\dfrac{-2x}{2\sqrt{1-x^2}}}{1-x^2}$

$\longrightarrow y_2=\dfrac{mx\cos(m\sin^{-1}x)-m^2\sqrt{1-x^2}\sin(m\sin^{-1}x)}{(1-x^2)\sqrt{1-x^2}}$

$\longrightarrow(1-x^2)y_2=\dfrac{mx\cos(m\sin^{-1}x)-m^2\sqrt{1-x^2}\sin(m\sin^{-1}x)}{\sqrt{1-x^2}}$

$\longrightarrow(1-x^2)y_2=\dfrac{mx\cos(m\sin^{-1}x)}{\sqrt{1-x^2}}-m^2\sin(m\sin^{-1}x)$

From (1) and (2),

$\longrightarrow(1-x^2)y_2=xy_1-m^2y$

$\longrightarrow\underline{\underline{(1-x^2)y_2-xy_1+m^2y=0}}$

Hence Proved!

Answer 3

Answer:

Note that $y_1=\dfrac{dy}{dx}$ and $y_2=\dfrac{d^2y}{dx^2}.$

Given,

$\longrightarrow y=\sin(m\sin^{-1}x)$

$\longrightarrow m^2y=m^2\sin(m\sin^{-1}x)\quad\quad\dots(1)$

First derivative of $y$ is,

$\longrightarrow y_1=\dfrac{d}{dx}\left[\sin(m\sin^{-1}x)\right]$

$\longrightarrow y_1=\cos(m\sin^{-1}x)\cdot \dfrac{m}{\sqrt{1-x^2}}$

$\longrightarrow y_1=\dfrac{m\cos(m\sin^{-1}x)}{\sqrt{1-x^2}}$

$\longrightarrow xy_1=\dfrac{mx\cos(m\sin^{-1}x)}{\sqrt{1-x^2}}\quad\quad\dots(2)$

Second derivative of $y$ is,

$\longrightarrow y_2=\dfrac{d}{dx}\left[\dfrac{m\cos(m\sin^{-1}x)}{\sqrt{1-x^2}}\right]$

$\longrightarrow y_2=m\cdot\dfrac{-\sin(m\sin^{-1}x)\cdot\dfrac{m}{\sqrt{1-x^2}}\cdot\sqrt{1-x^2}-\cos(m\sin^{-1}x)\cdot\dfrac{-2x}{2\sqrt{1-x^2}}}{1-x^2}$

$\longrightarrow y_2=\dfrac{mx\cos(m\sin^{-1}x)-m^2\sqrt{1-x^2}\sin(m\sin^{-1}x)}{(1-x^2)\sqrt{1-x^2}}$

$\longrightarrow(1-x^2)y_2=\dfrac{mx\cos(m\sin^{-1}x)-m^2\sqrt{1-x^2}\sin(m\sin^{-1}x)}{\sqrt{1-x^2}}$

$\longrightarrow(1-x^2)y_2=\dfrac{mx\cos(m\sin^{-1}x)}{\sqrt{1-x^2}}-m^2\sin(m\sin^{-1}x)$

From (1) and (2),

$\longrightarrow(1-x^2)y_2=xy_1-m^2y$

$\longrightarrow\underline{\underline{(1-x^2)y_2-xy_1+m^2y=0}}$

Hence Proved!

Answer 4

Answer:

Note that $y_1=\dfrac{dy}{dx}$ and $y_2=\dfrac{d^2y}{dx^2}.$

Given,

$\longrightarrow y=\sin(m\sin^{-1}x)$

$\longrightarrow m^2y=m^2\sin(m\sin^{-1}x)\quad\quad\dots(1)$

First derivative of $y$ is,

$\longrightarrow y_1=\dfrac{d}{dx}\left[\sin(m\sin^{-1}x)\right]$

$\longrightarrow y_1=\cos(m\sin^{-1}x)\cdot \dfrac{m}{\sqrt{1-x^2}}$

$\longrightarrow y_1=\dfrac{m\cos(m\sin^{-1}x)}{\sqrt{1-x^2}}$

$\longrightarrow xy_1=\dfrac{mx\cos(m\sin^{-1}x)}{\sqrt{1-x^2}}\quad\quad\dots(2)$

Second derivative of $y$ is,

$\longrightarrow y_2=\dfrac{d}{dx}\left[\dfrac{m\cos(m\sin^{-1}x)}{\sqrt{1-x^2}}\right]$

$\longrightarrow y_2=m\cdot\dfrac{-\sin(m\sin^{-1}x)\cdot\dfrac{m}{\sqrt{1-x^2}}\cdot\sqrt{1-x^2}-\cos(m\sin^{-1}x)\cdot\dfrac{-2x}{2\sqrt{1-x^2}}}{1-x^2}$

$\longrightarrow y_2=\dfrac{mx\cos(m\sin^{-1}x)-m^2\sqrt{1-x^2}\sin(m\sin^{-1}x)}{(1-x^2)\sqrt{1-x^2}}$

$\longrightarrow(1-x^2)y_2=\dfrac{mx\cos(m\sin^{-1}x)-m^2\sqrt{1-x^2}\sin(m\sin^{-1}x)}{\sqrt{1-x^2}}$

$\longrightarrow(1-x^2)y_2=\dfrac{mx\cos(m\sin^{-1}x)}{\sqrt{1-x^2}}-m^2\sin(m\sin^{-1}x)$

From (1) and (2),

$\longrightarrow(1-x^2)y_2=xy_1-m^2y$

$\longrightarrow\underline{\underline{(1-x^2)y_2-xy_1+m^2y=0}}$

Hence Proved!