#### Provide solution for RD Sharma maths class 12 chapter Differential Equation exercise 21.2 question 16 subquestion (vi)

The required differential equation is

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

Hint:

This is the equation of hyperbola

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

Given:

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

Solution:

The equation of family of curves is

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

Where a and b are parameter

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

$\frac{2x}{a^{2}}-\frac{2y}{b^{2}}\frac{\mathrm{d} y}{\mathrm{d} x}=0 \qquad \qquad \dots(ii)$

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

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

Now, from (ii), we get

\begin{aligned} &\frac{2x}{a^{2}}=\frac{2y}{b^{2}}\frac{\mathrm{d} y}{\mathrm{d} x}\\ &\frac{b^{2}}{a^{2}}=\frac{y}{x}\frac{\mathrm{d} y}{\mathrm{d} x} \qquad \qquad \dots (iv) \end{aligned}

From (iii) and (iv), we get

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

The required differential equation is

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