#### Please Solve R.D.Sharma class 12 Chapter 21  Differential Equations Exercise Revision Exercise Question 10 Maths textbook Solution.

Answer:$xy{y}''+x\left ( {y}' \right )^{2}-y{y}'=0$

Hint: You must know about the equation of hyperbola

Given: Family of hyperbola having foci on x-axis and centre at origin

Solution: Hyperbola whose foci on y-axis

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$                       [Two constants, differentiate twice]

\begin{aligned} &\frac{d}{d x}\left[\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\right]=\frac{d}{d x}(1) \\ &\frac{1}{a^{2}}\left(\frac{d\left(x^{2}\right)}{d x}\right)-\frac{1}{b^{2}}\left(\frac{d\left(y^{2}\right)}{d x}\right)=0 \\ &\frac{1}{a^{2}}(2 x)-\frac{1}{b^{2}}\left(2 y \frac{d y}{d x}\right)=0 \\ &\frac{2 x}{a^{2}}-\frac{2 y}{b^{2}}\left(\frac{d y}{d x}\right)=0 \end{aligned}

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

Again differentiate,

\begin{aligned} &y^{\prime} \frac{\left[\left(y y^{\prime}\right) x-y\right]}{x^{2}}+\frac{y}{x} y^{\prime \prime}=0 . . \text { using product and division rule of differentiation }\\ &\left(y^{\prime} y^{\prime}+y y^{\prime \prime}\right) x-y y^{\prime}=0 . . \text { multiplying by } x^{2}\\ &\left(y^{\prime 2}+y y^{\prime \prime}\right) x-y y^{\prime}=0\\ &x y y^{\prime \prime}+x\left(y^{\prime}\right)^{2}-y y^{\prime}=0 \end{aligned}