Helen Lewis

2021-12-31

Ronnie Schechter

$⇒x\left(3{y}^{2}-xy-{x}^{2}\right)dy=-y\left({x}^{2}+xy-2{y}^{2}\right)dx$
$⇒\frac{dy}{dx}=\frac{-y\left({x}^{2}+xy-2{y}^{2}\right)}{x\left(3{y}^{2}-xy-{x}^{2}\right)}$
$=\frac{-{x}^{2}y-xy+2{y}^{3}}{3x{y}^{2}-{x}^{2}y-{x}^{3}}$
$=\frac{\frac{-{x}^{2}y-xy+2{y}^{3}}{{x}^{3}}}{\frac{3x{y}^{2}-{x}^{3}y-{x}^{2}}{{x}^{3}}}$
$⇒\frac{dy}{dx}=\frac{\left(-\frac{y}{x}-\frac{{y}^{2}}{{x}^{2}}+2\frac{{y}^{3}}{{x}^{3}}\right)}{\left(3\frac{{y}^{2}}{{x}^{2}}-\frac{y}{x}-1\right)}$
Which is in the form: $\frac{dy}{dx}=F\left(\frac{y}{x}\right)=F\left(\frac{y}{x}\right)$
i.e. the above differential eq is homogeneous equation
$⇒y=vx$
$⇒\frac{dy}{dx}=v+\frac{dv}{dx}$
$⇒v+\frac{dv}{dx}=\frac{\left(2{v}^{3}-{v}^{2}-v\right)}{3{v}^{2}-v-1}$
$⇒\frac{dv}{dx}=\frac{2{v}^{3}-{v}^{2}-v}{3{v}^{2}-v-1}-v$

censoratojk

Given differential equation:
$y\left({x}^{2}+xy-2{y}^{2}\right)dx+x\left(3{y}^{2}-xy-{x}^{2}\right)dy=0$
$⇒\frac{dy}{dx}=-\frac{y\left({x}^{2}+xy-2{y}^{2}\right)}{x\left(3{y}^{2}-xy-{x}^{2}\right)}$
$=-\left(\frac{\frac{y}{x}+{\left(\frac{y}{x}\right)}^{2}-2{\left(\frac{y}{x}\right)}^{3}}{3{\left(\frac{y}{x}\right)}^{2}-\frac{y}{x}-1}\right)$ (i)
Put $\frac{y}{x}=v$
$⇒y=vx$
$⇒\frac{dy}{dx}=v+x\frac{dv}{dx}$
The equation (i) becomes
$v+x\frac{dv}{dx}=-\frac{v+{v}^{2}-2{v}^{3}}{3{v}^{2}-v-1}$
$⇒x\frac{dv}{dx}=-\left(\frac{v+{v}^{2}-2{v}^{3}}{3{v}^{2}-v-1}\right)+v\right)$
$=-\left(\frac{v+{v}^{2}-2{v}^{3}+3{v}^{3}-{v}^{2}-v}{3{v}^{2}-v-1}\right)$
$x\frac{dv}{dx}=-\left(\frac{{v}^{3}}{3{v}^{2}-v-1}\right)$
$⇒\frac{3{v}^{2}-v-1}{{v}^{3}}dv=\frac{-dx}{x}$
$⇒\left(\frac{3}{v}-\frac{1}{{v}^{2}}-\frac{1}{{v}^{3}}\right)dv=-\frac{dx}{x}$
on integration, we get

Vasquez