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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 16 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 16 Maths Textbook Solution.

Answers (1)

Answer: \sqrt{x^{2}+y^{2}}=ce^{2\tan -1}\frac{y}{x},x\neq 0

Given:\left ( x+2y \right )dx-\left ( 2x-y \right )dy=0

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

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

Solution: We have,

\left ( x+2y \right )dx-\left ( 2x-y \right )dy=0

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

It is a homogeneous equation.

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

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

\begin{aligned} &\Rightarrow v+x \frac{d v}{d x}=\frac{x(1+2 v)}{x(2-v)} \\ &\Rightarrow x \frac{d v}{d x}=\frac{1+2 v}{2-v}-v \\ &\Rightarrow x \frac{d v}{d x}=\frac{1+v^{2}}{2-v} \end{aligned}

Separating the variables

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

\Rightarrow \int \frac{d v}{1+v^{2}}-\int \frac{v}{1+v^{2}}=\int \frac{d x}{x}[\therefore \text { Integrating }]\\

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

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

\Rightarrow 2 \tan ^{-1} v=(1+v)^{\frac{1}{2}} x c\\

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

\Rightarrow e^{2 \tan ^{-1}} \frac{y}{x}=\left\{\frac{x^{2}+y^{2}}{x}\right\}^{\frac{1}{2}} x c\\

\Rightarrow e^{2 \tan ^{-1}} \frac{y}{x}=\left(x^{2}+y^{2}\right)^{\frac{1}{2}} c\\

\Rightarrow \sqrt{x^{2}+y^{2}}=\frac{1}{c} e^{2 \tan ^{-1}} \frac{y}{x}\\

\Rightarrow \sqrt{x^{2}+y^{2}}=C e^{2 \tan ^{-1} \frac{y}{x}} ; \text { Where }=\frac{1}{c}\\

This is required solution.

Posted by

infoexpert21

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