#### Need solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.2 question 8

$\left [ 1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \right ]^{3}=r^{2}\left ( \frac{\mathrm{d}^{2}y }{\mathrm{d} x^{2}} \right )^{2}$

Hint:

Differentiating the given equation where r is a constant

Given:

$(x-a)^{2}+(y-b)^{2}=r^{2} \qquad \qquad \dots(i)$

Solution:

$(x-a)^{2}+(y-b)^{2}=r^{2}$

\begin{aligned} &2(x-a)+2(y-b)\frac{\mathrm{d}y }{\mathrm{d} x}=0 \qquad \qquad \dots(ii)\\ &1+(y-b)\frac{\mathrm{d}^{2}y }{\mathrm{d} x^{2}}+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}=0 \\ &(y-b)\frac{\mathrm{d}^{2}y }{\mathrm{d} x^{2}}=-\left ( 1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \right ) \qquad \qquad \dots(iii) \end{aligned}

From equation (ii)

\begin{aligned} &(x-a)=-(y-b)\frac{\mathrm{d} y}{\mathrm{d} x} \end{aligned}

Put this value in equation (i)

\begin{aligned} &\left [ -(y-b)\frac{\mathrm{d} y}{\mathrm{d} x} \right ]^{2}+(y-b)^{2}=r^{2} \qquad \qquad \dots (iv) \end{aligned}

Squaring equation (iii),

\begin{aligned} &(y-b)^{2} \left ( \frac{\mathrm{d}^{2}y }{\mathrm{d} x^{2}} \right )^{2}=\left ( 1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \right )^{2} \qquad \qquad \dots(v) \end{aligned}

Dividing equation (iv) by (v),

\begin{aligned} &\frac{\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+1}{\left ( \frac{\mathrm{d}^{2}y }{\mathrm{d} x^{2}} \right )^{2}}=\frac{r^{2}}{\left [ \left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}+1 \right ]} \end{aligned}

$\left [ 1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \right ]^{3}=r^{2}\left ( \frac{\mathrm{d}^{2}y }{\mathrm{d} x^{2}} \right )^{2}$