#### Provide Solution for RD Sharma Class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Question question 19

(a)

Hint:

We must know about the derivative of $x$

Given:

$y=\frac{a x+b}{x^{2}+c} \$

Explanation:

$y=\frac{a x+b}{x^{2}+c} \$

\begin{aligned} &\ &\left(x^{2}+c\right) y=a x+b \end{aligned}

Differentiate with respect to $x$

$2 x y+\left(x^{2}+c\right) \frac{d y}{d x}=a$

Again differentiate,

$2 y+2 x y_{1}+2 x y_{1}+\left(x^{2}+c\right) y_{2}=0 \\$

\begin{aligned} &2 y+4 x y_{1}+\left(x^{2}+c\right) y_{2}=0 \end{aligned}

Differentiate again with respect to $x$

$2 y_{1}+4 y_{1}+4 x y_{2}+\left(x^{2}+c\right) y_{3}+2 x y_{2}=0 \\$

\begin{aligned} &6 y_{1}+6 x y_{2}+\left(x^{2}+c\right) y_{3}=0 \end{aligned}

$6 y_{1}+6 x y_{2}+\left(\frac{-2 y-4 x y_{1}}{y_{2}}\right) y_{3}=0 \\$

$6 y_{1} y_{2}+6 x\left(y_{2}\right)^{2}-2 y-4 x y_{1} y_{3}=0 \\$

$3 y_{1} y_{2}+3 x\left(y_{2}\right)^{2}-y-2 x y_{1} y_{3}=0$

\begin{aligned} \\ &\left(2 x y_{1}+y\right) y_{3}=\left(y_{1}+x y_{2}\right) 3 y_{2} \end{aligned}