Question

plz help me friends in this ​

Answer 1

$\color{darkcyan} \underline{ \begin{array}{ || |l| || } \hline \color{magenta} \\ \hline \boxed{ \text{ \tt \: Solution:-} } \end{array}}$

$\text{ We have \( \rm y=\cos \left(m \sin ^{-1} x\right) \)}$

$\begin{aligned} \rm y_{1} & \rm=-\sin \left(m \sin ^{-1} x\right) \cdot \frac{m}{\sqrt{1-x^{2}}} \\ \\ \rm y_{1}^{2} & \rm=\sin ^{2}\left(m \sin ^{-1} x\right) \frac{m^{2}}{\left(1-x^{2}\right)} \\ \\ \text { This implies } \rm \quad\left(1-x^{2}\right) y_{1}^{2} & \rm=m^{2} \sin ^{2}\left(m \sin ^{-1} x\right)=m^{2}\left[1-\cos ^{2}\left(m \sin ^{-1} x\right)\right] \\ \\ \text{ That is,} \rm\left(1-x^{2}\right) y_{1}^{2} & \rm=m^{2}\left(1-y^{2}\right) . \end{aligned}$

Again differentiating,

$\[ \begin{aligned} \rm \left(1-x^{2}\right) 2 y_{1} \frac{d y_{1}}{d x}+y_{1}{ }^{2}(-2 x) & \rm=m^{2}\left(-2 y \frac{d y}{d x}\right) \\ \\ \rm \left(1-x^{2}\right) 2 y_{1} y_{2}-2 x y_{1}{ }^{2} & \rm=-2 m^{2} y y_{1} \end{aligned} \]$

$\\ \rm\ \[ \left(1-x^{2}\right) y_{2}-x y_{1}=-m^{2} y \]$

Once again differentiating,

$\[ \begin{aligned} \rm \left(1-x^{2}\right) \frac{d y_{2}}{d x}+y_{2}(-2 x)-\left[x \cdot \frac{d y_{1}}{d x}+y_{1} \cdot 1\right] & \rm=-m^{2} \frac{d y}{d x} \\ \\ \rm\left(1-x^{2}\right) y_{3}-2 x y_{2}-x y_{2}-y_{1} & \rm=-m^{2} y_{1} \\ \\ \rm \left(1-x^{2}\right) y_{3}-3 x y_{2}+\left(m^{2}-1\right) y_{1} & \color{red}=0 . \end{aligned} \]$

Answer 2

Step-by-step explanation:

$\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}$

$\text{ We have \( \rm y=\cos \left(m \sin ^{-1} x\right) \)}$

$\begin{gathered} \begin{aligned} \rm y_{1} & \rm=-\sin \left(m \sin ^{-1} x\right) \cdot \frac{m}{\sqrt{1-x^{2}}} \\ \\ \rm y_{1}^{2} & \rm=\sin ^{2}\left(m \sin ^{-1} x\right) \frac{m^{2}}{\left(1-x^{2}\right)} \\ \\ \text { This implies } \rm \quad\left(1-x^{2}\right) y_{1}^{2} & \rm=m^{2} \sin ^{2}\left(m \sin ^{-1} x\right)=m^{2}\left[1-\cos ^{2}\left(m \sin ^{-1} x\right)\right] \\ \\ \text{ That is,} \rm\left(1-x^{2}\right) y_{1}^{2} & \rm=m^{2}\left(1-y^{2}\right) . \end{aligned} \end{gathered}$

Again differentiating,

$\begin{gathered} \begin{aligned} \rm \left(1-x^{2}\right) 2 y_{1} \frac{d y_{1}}{d x}+y_{1}{ }^{2}(-2 x) & \rm=m^{2}\left(-2 y \frac{d y}{d x}\right) \\ \\ \rm \left(1-x^{2}\right) 2 y_{1} y_{2}-2 x y_{1}{ }^{2} & \rm=-2 m^{2} y y_{1} \end{aligned} \end{gathered}$

$\begin{gathered} \\ \rm\ \left(1-x^{2}\right) y_{2}-x y_{1}=-m^{2} y \end{gathered}$

Once again differentiating,

$\begin{gathered} \begin{aligned} \rm \left(1-x^{2}\right) \frac{d y_{2}}{d x}+y_{2}(-2 x)-\left[x \cdot \frac{d y_{1}}{d x}+y_{1} \cdot 1\right] & \rm=-m^{2} \frac{d y}{d x} \\ \\ \rm\left(1-x^{2}\right) y_{3}-2 x y_{2}-x y_{2}-y_{1} & \rm=-m^{2} y_{1} \\ \\ \rm \left(1-x^{2}\right) y_{3}-3 x y_{2}+\left(m^{2}-1\right) y_{1} & \color{red}=0 . \end{aligned} \end{gathered}$