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Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 72 textbook solution.

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Answer :x^{2}-y^{2}=c x

Given : \frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}


Using variable separable method.


We know that the slope of the tangent at(x,y) of a Curve is  \frac{dy}{dx}


Given slope of tangent at (x,y) is \frac{x^{2}+y^{2}}{2 x y}

\therefore \frac{d y}{d x}=\frac{x^{2}+y^{2}}{2 x y}-(1)

\begin{aligned} &\text { Put } \frac{d y}{d x}=f(x, y) \Rightarrow F(x, y)=\frac{x^{2}+y^{2}}{2 x y} \\ &\text { Finding } f(\lambda x, \lambda y)=\frac{(\lambda x)^{2}+(\lambda y)^{2}}{2(\lambda x)(\lambda y)} \\ \end{aligned}

                                        \begin{aligned} &=\frac{\lambda^{2} x^{2}+\lambda^{2} y^{2}}{2 \lambda^{2} x y} \end{aligned}

                                       \begin{aligned} &=\frac{\lambda^{2}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)}{\lambda^{2} \cdot 2 \mathrm{xy}} \\ \end{aligned}

                                      \begin{aligned} &=\frac{\mathrm{x}^{2}+\mathrm{y}^{2}}{2 \mathrm{xy}} \\ \end{aligned}

                                      \begin{aligned} &=\mathrm{F}(\mathrm{x}, \mathrm{y}) \\ \end{aligned}

\begin{aligned} &\text { So, } \mathrm{F}(\lambda \mathrm{x}, \lambda \mathrm{y})=\mathrm{F}(\mathrm{x}, \mathrm{y}) \\ \end{aligned}

                              \begin{aligned} &=\lambda^{\circ} \mathrm{F}(\mathrm{x}, \mathrm{y}) \end{aligned}

So \mathrm{F}(\lambda \mathrm{x}, \lambda \mathrm{y}) is homogeneous function of degree 0 

\therefore Given equation is homogeneous differential equation

Now Put y =vx in (1) 

Differentiate with respect to x

\begin{aligned} &\frac{d y}{d x}=x \frac{d v}{d x}+v\\ &\text { Putting } \frac{\mathrm{dy}}{\mathrm{dx}} \text { value in }(1) \text { and } \mathrm{y}=\mathrm{vx} \end{aligned}

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

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

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

Integrating both sides,

\begin{aligned} &\int \frac{2 \mathrm{v}}{\mathrm{v}^{2}-1} \mathrm{dv}=-\int \frac{\mathrm{dx}}{\mathrm{x}} \\ &\Rightarrow \int \frac{\partial \mathrm{v}}{\mathrm{v}^{2}-1} \mathrm{~d} \mathrm{v}=-\int \frac{\mathrm{dx}}{\mathrm{x}} \\ &\Rightarrow \int \frac{\partial \mathrm{v}}{\mathrm{v}^{2}-1} \mathrm{dv}=-\log |\mathrm{x}|+\mathrm{c} \end{aligned}

Solving \int \frac{2v}{v^{2}-1}dv

Put v^{2}-1=t

Differentiate with respect to v

\begin{gathered} 2 v d v=d t \\ d v=\frac{d t}{2 v} \\ =\int \frac{2 v}{t} \cdot \frac{d t}{2 v} \\ =\int \frac{d t}{t}=\log 1+1 \end{gathered}

Putting back

\begin{aligned} &\mathrm{t}=\mathrm{v}^{2}-1 \\ &\Rightarrow \log \left|\mathrm{v}^{2}-1\right| \end{aligned}

\therefore By (2)

\begin{aligned} &\log \left|\mathrm{v}^{2}-1\right|=-\log |\mathrm{x}|+\mathrm{c}_{1}\\ &\text { Putting }\\ &\mathrm{vx}=\mathrm{y} \text { or } \end{aligned}

\begin{aligned} &\mathrm{v}=\frac{\mathrm{y}}{\mathrm{x}} \\ &\log \left|\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-1\right|=-\log |\mathrm{x}|+\mathrm{C}_{1} \\ &\Rightarrow \log \left|\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-1\right|+\log |\mathrm{x}| \\ &=\mathrm{C}_{1} \end{aligned}

\begin{aligned} &\Rightarrow \log \left|\left[\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-1\right] \dot{\mathrm{x}}\right| \\ &=\mathrm{C}_{1} \\ &\Rightarrow \log \left|\left[\frac{\mathrm{y}^{2}}{\mathrm{x}^{2}}-1\right] \mathrm{x}\right|=\mathrm{C}_{1} \end{aligned}

Putting C_{1}=\log\; C_{1}

\log \left|\left(\frac{\mathrm{y}^{2}}{\mathrm{x}^{2}}-1\right) \mathrm{x}\right|=\log \mathrm{C}_{1}

Remove log of both sides.

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

\begin{aligned} \\\Rightarrow & \frac{x y^{2}}{x^{2}}-x=C_{1} \\ \end{aligned}

\begin{aligned} \\\Rightarrow \frac{y^{2}}{x}-x=C_{1} \end{aligned}

\begin{aligned} &\Rightarrow \mathrm{y}^{2}-\mathrm{x}^{2}=\mathrm{x} \mathrm{C}_{1} \\ \\&\Rightarrow \mathrm{x}^{2}-\mathrm{y}^{2}=-\mathrm{C}_{1} \mathrm{x} \\ \\&\text { Put } \mathrm{C}=-\mathrm{C}_{1} \\ \\&\Rightarrow \mathrm{x}^{2}-\mathrm{y}^{2}=\mathrm{cx} \end{aligned}

Hence Proved


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