#### Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 24 maths

Proved

Hint:

You must know about the derivative of sin function and logarithm function

Given:

$x=\sin \left(\frac{1}{a} \log y\right) \text { show that }\left(1-x^{2}\right) y_{2}-x y_{1}-a^{2} y=0$

Solution:

\begin{aligned} &x=\sin \left(\frac{1}{a} \log y\right)\\ &\log y=a \sin ^{-1} x\\ &y=e^{a \sin ^{-1} x}\; \; \; \; \; \; \; .....(1) \end{aligned}

\begin{aligned} &\text { Let } t=a \sin ^{-1} x\\ &\frac{d t}{d x}=\frac{a}{\sqrt{1-x^{2}}} \quad\left[\frac{d}{d x} \sin ^{-1} x=\frac{1}{\sqrt{1-x^{2}}}\right]\\ &\text { and } y=e^{t}\\ &\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}\\ &\frac{d y}{d x}=e^{t} \frac{a}{\sqrt{1-x^{2}}}=\frac{a e^{a \sin ^{-1} x}}{\sqrt{1-x^{2}}}\; \; \; \; \; \; \; \; .......(2) \end{aligned}

\begin{aligned} &\text { Again differentiating both sides }\\ &\frac{d^{2} y}{d x^{2}}=a e^{a \sin ^{-1} x} \frac{d}{d x}\left(\frac{1}{\sqrt{1-x^{2}}}\right)+\frac{a}{\sqrt{1-x^{2}}} \frac{d}{d x} e^{a \sin ^{-1} x} \end{aligned}

\begin{aligned} &\text { Using chain rule and equation }\\ &\frac{d^{2} y}{d x^{2}}=\frac{-a e^{a \sin ^{-1} x}}{2\left(1-x^{2}\right) \sqrt{1-x^{2}}}(-2 x)+\frac{a^{2} e^{a \sin ^{-1} x}}{\left(1-x^{2}\right)}\\ &\frac{d^{2} y}{d x^{2}}=\frac{x a e^{a \sin ^{-1} x}}{\left(1-x^{2}\right) \sqrt{1-x^{2}}}+\frac{a^{2} e^{a \sin ^{-1} x}}{\left(1-x^{2}\right)}\\ &\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=a^{2} e^{a \sin ^{-1} x}+\frac{x a e^{a \sin ^{-1} x}}{\sqrt{1-x^{2}}} \end{aligned}

\begin{aligned} &\text { Using equation }(1) \text { and }(2)\\ &\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=a^{2} y+x \frac{d y}{d x}\\ &\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-a^{2} y-x \frac{d y}{d x}=0\\ &\left(1-x^{2}\right) y_{2}-x y_{1}-a^{2} y=0 \end{aligned}