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Provide Solution For  R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 21 Maths Textbook Solution.

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Given: \left [ x\sqrt{x^{2}+y^{2}}-y^{2} \right ]dx+xydy=0

To solve: we have to solve the given differential equation

Hint: Put y=vx and \frac{dy}{dx}=v+x\frac{dv}{dx}

Solution: \left [ x\sqrt{x^{2}+y^{2}}-y^{2} \right ]dx+xydy=0

\Rightarrow \frac{dy}{dx}=\frac{\left [ y^{2}-x\sqrt{x^{2}+y^{2}} \right ]}{xy}

It is a homogeneous equation.

Putting  y=vx and \frac{dy}{dx}=v+x\frac{dv}{dx}

So,v+x \frac{d v}{d x}=\frac{v^{2} x^{2}-x \sqrt{x^{2}+v^{2}} x^{2}}{v x^{2}}

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

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

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

\Rightarrow \int \frac{v}{\sqrt{\left(1+v^{2}\right)}} d v=-\frac{d x}{x} \\                                            (seperating \: the \: variables\: and\: intersecting \: both \: sides)

\Rightarrow \frac{1}{2} \int \frac{2 v}{\sqrt{\left(1+v^{2}\right)}} d v=-\int \frac{d x}{x}



\Rightarrow \frac{1}{2} \int \frac{1}{\sqrt{t}} d t=-\log x+\log c \\


                                                                              \Rightarrow \sqrt{1+v^{2}}=\log \left(\frac{c}{x}\right)\left[\therefore t=1+v^{2}\right] \\

                                                                                \Rightarrow \frac{\sqrt{x^{2}+y^{2}}}{x} =\log \left(\frac{c}{x}\right)\left[\therefore v=\frac{y}{x}\right] \\

\Rightarrow \sqrt{x^{2}+y^{2}}=x \log \left(\frac{c}{x}\right)

This is required solution.

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