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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.

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

Answers (1)

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.

Posted by

infoexpert21

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