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Need Solution for R.D .Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 35 Maths Textbook Solution.

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Answer: Cy=log|\frac{y}{x}|-1

Given:ydx+\left \{ xlog\frac{y}{x} \right \}dy-2xdy=0

To solve: we have to solve differential equation

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

Solution: we have,

ydx+\left \{ xlog\frac{y}{x} \right \}dy-2xdy=0

Divided by dx both side we get

\begin{aligned} &\Rightarrow y+x \log \left(\frac{y}{x}\right) \frac{d y}{d x}-2 x \frac{d y}{d x}=0 \\ &\Rightarrow \frac{d y}{d x}=\frac{y}{2 x-x \log \left(\frac{y}{x}\right)} \end{aligned}

It is homogeneous equation

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

So,  x \frac{d y}{d x}+v=\frac{v x}{2 x-x \log \left(\frac{y}{x}\right)} \\

\Rightarrow x \frac{d y}{d x}=\frac{v x}{2 x-\log v}-v \\

\Rightarrow x \frac{d y}{d x}=\frac{v x}{2 x-\log v}-v \\

\Rightarrow x \frac{d y}{d x}=\frac{v-2 v+2 v x \log v}{2-\log v}

Separating the variables and integrating both side we get

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

Let t=logv-1

\Rightarrow dt=\frac{1}{v}-1

Ow equation becomes

\Rightarrow \frac{d t}{t}-\log v=\log x+\log c\\

\Rightarrow \log t-\log v=\log x+\log c\\

\Rightarrow \log |\log v-1|-\log v=\log x+\log c \quad[\therefore t=\log v-1]\\

\Rightarrow \log (\log v-1)=\log x+\log v+\log c\\

\Rightarrow \log (\log v-1)=\log c v x \quad \text { (using logrithm property) }\\

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

\Rightarrow \log \frac{y}{x}-1=c y\\

\Rightarrow c y=\log \left|\frac{y}{x}\right|-1

This is required solution.

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