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

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

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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.

 

Posted by

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