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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21 .9 Question 20 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21 .9 Question 20 Maths Textbook Solution.

Answers (1)

Answer: y=c e^{\tan ^{-1}\left(\frac{y}{x}\right)}

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

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

Solution: we have

y^{2}dx+\left ( x^{2}-xy+y^{2} \right )dy=0

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

It is homogeneous equation.

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

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

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

\begin{aligned} &\Rightarrow x \frac{d v}{d x}=\frac{-v^{2}-v+v^{2}-v^{3}}{1-v+v^{2}} \\ &\Rightarrow x \frac{d v}{d x}=\frac{-v-v^{3}}{1-v+v^{2}} \\ &\Rightarrow x \frac{d v}{d x}=\frac{-v\left(v^{2}+1\right)}{v^{2}-v+1} \end{aligned}

Separating the variables we have,

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

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

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

\Rightarrow-\operatorname{logv}+\tan ^{-1} v=\log x=\log c \\

\Rightarrow-\log \left(\frac{y}{x}\right)+\tan ^{-1}\left(\frac{y}{x}\right)=\log x k \\                                                                    \left [ \therefore v=\frac{y}{x} \right ]

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

\Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)=\log x k-\log \left(\frac{x}{y}\right) \\

\Rightarrow \tan ^{-1}\left(\frac{y}{x}\right)=\log \left(\frac{x k y}{x}\right) \\

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

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

\Rightarrow \mathrm{y}=c e^{\tan ^{-1}\left(\frac{y}{x}\right)} \quad \text { Where } c=\frac{1}{k}

This is required solution.

Posted by

infoexpert21

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