Solve differential equation y(2x-2+xy+1)dx+(x-y)dy=0

Daniaal Sanchez 2020-12-17 Answered
Solve differential equation $y\left(2x-2+xy+1\right)dx+\left(x-y\right)dy=0$
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Fatema Sutton
$y\left(2x-2+xy+1\right)dx+\left(x-y\right)dy=0$
$y\left(2x-2+xy+1\right)+\left(x-y\right)\frac{dy}{dx}=0$
$y\left(2x-2+xy+1\right)+\left(x-y\right){y}^{\prime }=0$
${y}^{\prime }=-\frac{y\left(2{x}^{2}-xy+1\right)}{x-y}$
$y=xv⇒\frac{dy}{dx}=v+x\frac{dy}{dx}$
${y}^{\prime }=-\frac{y\left(2{x}^{2}-xy+1\right)}{x-y}⇒\left(x-y\right){y}^{\prime }+y\left(2{x}^{2}-xy+1\right)=0$
$\left(x-y\right)bigg\left(v+x\frac{dv}{dx}bigg\right)+xv\left(2{x}^{2}-xy+1\right)=0$
$\frac{dv}{dx}=-\frac{-{v}^{2}-{x}^{2}{v}^{2}+2v+2{x}^{2}v}{x\left(v-1\right)}$
$\frac{dv}{dx}=-\frac{\left({x}^{2}+1\right)\left(v-2\right)v}{x\left(v-1\right)}$
$\frac{1}{v}bigg\left(\frac{v-1}{v-2}bigg\right)\frac{dv}{dx}-\frac{{x}^{2}+1}{x}$
Integrating both sides with respect to x
$\frac{1}{v}bigg\left(\frac{v-1}{v-2}bigg\right)\left\{dv\right\}=-\int \frac{\left({x}^{2}+1\right)}{x}dx$
Jeffrey Jordon