#### Explain solution RD Sharma class 12 chapter Differential Equation exercise 21.2 question 16 subquestion (iv) maths

The required differential equation is

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

Hint:

$\text { Putting } \sqrt{1-x^{2}} \text { instead of } (y-b) \text { in }2x+2(y-b)\frac{\mathrm{d} y}{\mathrm{d} x}=0$

Given:

$x^{2}+(y-b)^{2}=1$

Solution:

The equation of family of curves is

$x^{2}+(y-b)^{2}=1 \qquad \qquad \dots(i)$

Where b is a parameter

Differentiating equation (i) with respect to x, we get

\begin{aligned} &2x+2(y-b)\frac{\mathrm{d} y}{\mathrm{d} x}=0 \\ &2x+2\sqrt{1-x^{2}}\frac{\mathrm{d} y}{\mathrm{d} x}=0 \qquad \qquad [Using (i)]\\ &x=-\sqrt{1-x^{2}}\frac{\mathrm{d} y}{\mathrm{d} x}\\ &x^{2}=(1-x^{2})\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}\\ &x^{2}=\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2}-x^{2}\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )^{2} \end{aligned}

The required equation is

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