1. ## Bernulli equations

1. $ydx+(x-\frac{1}{2}x^{3}y)dy=0
$

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2.x=(x^{2}-2y+1)y'

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2. Originally Posted by sinichko
1. $ydx+(x-\frac{1}{2}x^{3}y)dy=0
$

$

2.x=(x^{2}-2y+1)y'

$
$y\,dx + \left(x - \frac{1}{2}\,x^3y\right)dy = 0$

$\left(x - \frac{1}{2}\,x^3y\right)\frac{dy}{dx} + y = 0$

$\left(\frac{2x - x^3y}{2}\right)\frac{dy}{dx} + y = 0$

$\left(\frac{2x - x^3y}{2}\right)\frac{dy}{dx} = -y$

$\frac{dy}{dx} = \frac{2y}{x^3y - 2x}$

$\frac{dx}{dy} = \frac{x^3y - 2x}{2y}$

$\frac{dx}{dy} = \frac{x^3y}{2y} - \frac{2x}{2y}$

$\frac{dx}{dy} = \frac{1}{2}\,x^3 - \frac{1}{y}\,x$

$\frac{dx}{dy} + \frac{1}{y}\,x = \frac{1}{2}\,x^3$.

Now this is a Bernoulli Equation, so make the substitution $v = x^2$, so $x = v^{\frac{1}{2}}$.

Therefore $\frac{dx}{dy} = \frac{dx}{dv}\,\frac{dv}{dy}$

$\frac{dx}{dy} = \frac{1}{2}\,v^{-\frac{1}{2}}\,\frac{dv}{dy}$.

Substituting into the DE gives:

$\frac{dx}{dy} + \frac{1}{y}\,x = \frac{1}{2}\,x^3$

$\frac{1}{2}\,v^{-\frac{1}{2}}\,\frac{dv}{dy} + \frac{1}{y}\,v^{\frac{1}{2}} = \frac{1}{2}\,v^{\frac{3}{2}}$

$\frac{dv}{dy} + \frac{2}{y}\,v = v^2$.

This is now first order linear. So the integrating factor is $e^{\frac{2}{y}\,dy} = e^{2\ln{y}} = e^{\ln{y^2}} = y^2$.

Multiplying through by the integrating factor:

$\frac{dv}{dy} + \frac{2}{y}\,v = v^2$

$y^2\,\frac{dv}{dy} + 2y\,v = y^2v^2$

$\frac{d}{dy}\left(v\,y^2\right) = y^2v^2$

$v\,y^2 = \int{y^2v^2\,dy}$

$v\,y^2 = \frac{v^2y^3}{3} - \int{\frac{2v\,y^3}{3}\,dy}$

$v\,y^2 = \frac{v^2y^3}{3} - \frac{2}{3}\int{v\,y^3\,dy}$

$v\,y^2 = \frac{v^2y^3}{3} - \frac{2}{3}\left(\frac{v\,y^4}{4} - \int{\frac{y^4}{4}\,dy}\right)$

$v\,y^2 = \frac{v^2y^3}{3} - \frac{v\,y^4}{6} + \frac{1}{6}\int{y^4\,dy}$

$v\,y^2 = \frac{v^2y^3}{3} - \frac{v\,y^4}{6} + \frac{y^5}{30} + C$.

And since $v = x^2$

$x^2y^2 = \frac{x^4y^3}{3} - \frac{x^2y^4}{6} + \frac{y^5}{30} + C$

$x^2y^2 - \frac{x^4y^3}{3} + \frac{x^2y^4}{6} - \frac{y^5}{30} = C$.

Edit: Actually it looks like I've done something wrong... I'll keep checking...