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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.

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

Answers (1)

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}

So,v+x\frac{dv}{dx}=\frac{vx}{x}-\sqrt{\frac{v^{2}x^{2}}{x^{2}}-1}

\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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