#### Please solve RD Sharma class 12 chapter Differential Equations exercise 21.11 question 29 maths textbook solution

Given: Slope of tangent  $=\frac{x^{2}+y^{2}}{2 x y}$.

To prove: The curves which slope at any point $(x, y) \text { is } \frac{x^{2}+y^{2}}{2 x y}$  are in rectangular hyperbola.

Hint: The homogeneous differential equation put $y=v x \frac{d y}{d x}=v+x \frac{d v}{d x}$

Solution: Shape of tangent  $=\frac{x^{2}+y^{2}}{2 x y}$

$=\frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}$

It is a homogeneous equation

Out $y=vx$

Then,

\begin{aligned} &=v+x \frac{d v}{d x}=\frac{x^{2}+v^{2} x^{2}}{2 v x^{2}} \\\\ &=v+x \frac{d v}{d x}=\frac{x^{2}\left(1+v^{2}\right)}{2 v x^{2}} \\\\ &=v+x \frac{d v}{d x}=\frac{1+v^{2}}{2 v} \end{aligned}

\begin{aligned} &=x \frac{d v}{d x}=\frac{1+v^{2}}{2 v}-v \\\\ &=x \frac{d v}{d x}=\frac{1+v^{2}-2 v^{2}}{2 v} \\\\ &=x \frac{d v}{d x}=\frac{1-v^{2}}{2 v} \end{aligned}

$=\frac{2 v}{1-v^{2}}=\frac{d x}{x}$

$=-\left(\frac{2 v}{v^{2}-1}\right)=\frac{d x}{x}$

$=\frac{2 v}{v^{2}-1}=-\frac{d x}{x}$        [Separating variables]

Integrating on both sides

\begin{aligned} &=\int \frac{2 v}{1-v^{2}} d v=-\int \frac{1}{x} d x \\\\ &=\int \frac{1}{t} d t=-\int \frac{1}{x} d x \quad\left[1-v^{2}=t, 2 v d v=d t\right] \end{aligned}

\begin{aligned} &=\log t=-\log x+\log C \\\\ &=\log \left(1-v^{2}\right)=-\log x+\log C \\\\ &=-\log x=\log \left(1-v^{2}\right)-\log C \\\\ &=-\log x=\log \frac{\left(1-v^{2}\right)}{c} \end{aligned}

\begin{aligned} &=\log x^{-1}=\log \frac{\left(1-v^{2}\right)}{c} \\\\ &=\log \frac{1}{x}=\log \frac{\left(1-v^{2}\right)}{c} \end{aligned}

Using antilogarithm

\begin{aligned} &=\frac{1}{x}=\frac{1-v^{2}}{c} \\\\ &=1-v^{2}=\frac{c}{x} \\\\ &=1-\left(\frac{y}{x}\right)^{2}=\frac{c}{x} \\\\ &=1-\frac{y^{2}}{x^{2}}=\frac{c}{x} \end{aligned}

\begin{aligned} &=\frac{x^{2}-y^{2}}{x^{2}}=\frac{c}{x} \\\\ &=x^{2}-y^{2}=\frac{c x^{2}}{x} \\\\ &=x^{2}-y^{2}=C x \end{aligned}

Hence, $x^{2}-y^{2}=C x$ is the equation for a rectangular hyperbola.