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  • Please Solve R.D.Sharma class 12 Chapter 21 Differential Equations Exercise 21.9 Question 7 Maths textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 21  Differential Equations Exercise 21.9 Question 7 Maths textbook Solution.

Answers (1)

Answer: x=c\left ( x^{2}+y^{2} \right )

Given:2xy \frac{dy}{dx}= x^{2}+y^{2}

To find: we have to find the solution of given differential equation.

Hint: IN homogeneous differential equation

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

Solution: We have

2xy \frac{dy}{dx}= x^{2}+y^{2}

\Rightarrow \frac{dy}{dx}= \frac{x^{2}+y^{2}}{2xy}

It is homogeneous equation

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

So,v+x\frac{dv}{dx}=\frac{x^{2}+v^{2}x^{2}}{2xvx}

\Rightarrow x\frac{dv}{dx}=\frac{1+v^{2}}{2v}-v

                                                                \Rightarrow x\frac{dv}{dx}=\frac{1+v^{2}-2v^{2}}{2v}

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

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

\Rightarrow \int \frac{-2 v}{1-v^{2}} d v=-\int \frac{d x}{x} \\                                        [Integrating\; on \; both\; side]

\Rightarrow \log \left|1-v^{2}\right|=-\log x+\log c \mid \\

\Rightarrow 1-v^{2}=\frac{c}{x}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \left[\therefore \log c-\log x=\log \frac{c}{x}\right] \\

\Rightarrow x\left(1-\frac{y^{2}}{x^{2}}\right)=C \; \; \; \; \; \; \; \; \quad\left[\therefore \text { put } v=\frac{y}{x}\right] \\

\Rightarrow x\left(\frac{x^{2}-y^{2}}{x^{2}}\right)=C \\

\Rightarrow x^{2}-y^{2}=c x                        which is required solution.

 

Posted by

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