Question
Solve the differential equation $\frac{ d y}{ d x}+\frac{x-2 y}{2 x-y}=0$
✓
Answer
$\frac{ d y}{ d x}++\frac{x-2 y}{2 x-y}=0\ldots(i)$
Put $y=v x$$\ldots(ii)$
Differentiating w.r.t. $x$, we get
$\frac{ d y}{ d x}= v +x \frac{ dv }{ d x}\ldots(iii)$
Substituting (ii) and (iii) in (i), we get
$ v +x \frac{ dv }{ d x}+\frac{x-2 v x}{2 x- v x}=0$
$\therefore x \frac{ dv }{ d x}+ v +\frac{1-2 v }{2- v }=0$
$\therefore x \frac{ dv }{ d x}+\frac{2 v - v ^2+1-2 v }{2- v }=0$
$\therefore x \frac{ dv }{ d x}+\frac{1- v ^2}{2- v }=0$
$\therefore x \frac{ dv }{ d x}=-\frac{1- v ^2}{2- v }$
$\therefore x \frac{ dv }{ d x}=-\frac{ v ^2-1}{ v -2}$
$\therefore \frac{ v -2}{ v ^2-1} dv =\frac{- dv }{x} $
Integrating on both sides, we get
$ \int \frac{ v -2}{ v ^2-1} dv =-\int \frac{ d x}{x}$
$\therefore \int \frac{ v }{ v ^2-1} dv -\int \frac{2}{ v ^2-1} dv =-\int \frac{ d x}{x}$
$\therefore \frac{1}{2} \int \frac{2 v }{ v ^2-1} dv -\int \frac{2}{ v ^2-1} dv =-\int \frac{ d x}{x}$
$\therefore \frac{1}{2} \log \left| v ^2-1\right|-2 \times \frac{1}{2} \log \left|\frac{ v -1}{ v +1}\right|=-\log | x |+\log | c |$
$\therefore \frac{1}{2} \log \left|\left(\frac{y}{x}\right)^2-1\right|-\log \left|\frac{\frac{y}{x}-1}{\frac{y}{x}+1}\right|=-\log | x |+\log | c |$
$\therefore \frac{1}{2} \log \left|\frac{y^2-x^2}{x}\right|-\log \left|\frac{y-x}{y+x}\right|=-\log | x |+\log | c |$
$\therefore \frac{1}{2} \log \left|y^2-x^2\right|-\frac{1}{2} \log \left|x^2\right|+\log \left|\frac{y+x}{y-x}\right|=-\log | x |+\log | c |$
$\therefore \frac{1}{2} \log \left|x^2-y^2\right|-\log |x|+\log \left|\frac{x+y}{x-y}\right|=-\log | x |+\log | c |$
$\therefore \log \left|+\frac{1}{2} \log \right| x^2-y^2|=\log | $
Put $y=v x$$\ldots(ii)$
Differentiating w.r.t. $x$, we get
$\frac{ d y}{ d x}= v +x \frac{ dv }{ d x}\ldots(iii)$
Substituting (ii) and (iii) in (i), we get
$ v +x \frac{ dv }{ d x}+\frac{x-2 v x}{2 x- v x}=0$
$\therefore x \frac{ dv }{ d x}+ v +\frac{1-2 v }{2- v }=0$
$\therefore x \frac{ dv }{ d x}+\frac{2 v - v ^2+1-2 v }{2- v }=0$
$\therefore x \frac{ dv }{ d x}+\frac{1- v ^2}{2- v }=0$
$\therefore x \frac{ dv }{ d x}=-\frac{1- v ^2}{2- v }$
$\therefore x \frac{ dv }{ d x}=-\frac{ v ^2-1}{ v -2}$
$\therefore \frac{ v -2}{ v ^2-1} dv =\frac{- dv }{x} $
Integrating on both sides, we get
$ \int \frac{ v -2}{ v ^2-1} dv =-\int \frac{ d x}{x}$
$\therefore \int \frac{ v }{ v ^2-1} dv -\int \frac{2}{ v ^2-1} dv =-\int \frac{ d x}{x}$
$\therefore \frac{1}{2} \int \frac{2 v }{ v ^2-1} dv -\int \frac{2}{ v ^2-1} dv =-\int \frac{ d x}{x}$
$\therefore \frac{1}{2} \log \left| v ^2-1\right|-2 \times \frac{1}{2} \log \left|\frac{ v -1}{ v +1}\right|=-\log | x |+\log | c |$
$\therefore \frac{1}{2} \log \left|\left(\frac{y}{x}\right)^2-1\right|-\log \left|\frac{\frac{y}{x}-1}{\frac{y}{x}+1}\right|=-\log | x |+\log | c |$
$\therefore \frac{1}{2} \log \left|\frac{y^2-x^2}{x}\right|-\log \left|\frac{y-x}{y+x}\right|=-\log | x |+\log | c |$
$\therefore \frac{1}{2} \log \left|y^2-x^2\right|-\frac{1}{2} \log \left|x^2\right|+\log \left|\frac{y+x}{y-x}\right|=-\log | x |+\log | c |$
$\therefore \frac{1}{2} \log \left|x^2-y^2\right|-\log |x|+\log \left|\frac{x+y}{x-y}\right|=-\log | x |+\log | c |$
$\therefore \log \left|+\frac{1}{2} \log \right| x^2-y^2|=\log | $
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