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Provide Solution For  R.D. Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 18 Maths Textbook Solution.

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Answer: log \left ( \frac{y}{x} \right )=cx

Given:\frac{dy}{dx}=\frac{y}{x}\left \{ logy -log x+1 \right \}

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

Solution:\frac{dy}{dx}=\frac{y}{x}\left \{ logy -log x+1 \right \}

\Rightarrow \frac{d y}{d x}=\frac{y}{x}\left\{\log \frac{y}{x}-1\right\}\left[\therefore \log y-\log x=\log \frac{y}{x}\right]

It is homogeneous equation.

\begin{aligned} &\text { Put } y=v x \text { and } \frac{d y}{d x}=v+x \frac{d v}{d x} \\ &\text { So, } v+x \frac{d v}{d x}=\frac{v x}{x}\left\{\log \frac{v x}{x}+1\right\} \\ &\Rightarrow x \frac{d v}{d x}=v \log v \end{aligned}

Separating the variables we have,

\Rightarrow \int \frac{1}{v l o g v} d v=\int \frac{d x}{d x} \\

\therefore \int \frac{1}{v \log v} d v \\

\text { Put } \log v=t \\

\Rightarrow \frac{1}{v} d v=d t \\

\therefore \int \frac{1}{t} d t=\log t+c \\

\Rightarrow \log |\log v|+c \\

\therefore \log |\log |=\log |x|+\log c \\

\Rightarrow \operatorname{logv}=x c \\

\Rightarrow \log \frac{y}{x}=C x                                                                                                                                                                            \left [ \therefore v=\frac{y}{x} \right ]

This is required solution.

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