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Provide Solution For R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 17 Maths Textbook Solution.

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Answer: y+\sqrt{y^{2}-x^{2}}=c

Given:   \frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}

To solve: We have to solve the given differential equation.

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

Solution: \frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}

It is a homogeneous equation

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


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

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

\Rightarrow \int \frac{d v}{\sqrt{v^{2}-1}}=-\int \frac{d x}{x} \\

\Rightarrow \log \left|v+\sqrt{v^{2}-1}\right|=-\log x+\log c \\                                              \left[\therefore \int \frac{d v}{\sqrt{v^{2}-1}}=\log \left|v+\sqrt{v^{2}} 1\right|\right]                                          

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

\Rightarrow \frac{y}{x}+\sqrt{\frac{y^{2}}{x^{2}}-1}=\frac{c}{x}\left[\therefore v=\frac{y}{x}\right] \\

\Rightarrow \frac{y}{x}+\sqrt{\frac{y^{2}-x^{2}}{x}}=\frac{c}{x} \\

\Rightarrow \mathrm{y}+\sqrt{y^{2}-x^{2}}=c

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