Question

dy/dx + y/x = y^{3 solve in Bernoulli equation}

Answer 1

$\frac{dx}{dy} + \frac{y}{x} = y$

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

supposed

$\frac{1}{y^2} = v$ $\frac{-2}{y^3}.\frac{dy}{dx}=\frac{dv}{dx}$

$\frac{-1}{2}.\frac{dv}{dx}+ \frac{v}{x}=1$ $\frac{dv}{dx} \frac{-2}{x} = -2$

$\frac{dv}{dx} +$ $ev =Q$ for Bernoulli equation , $p=$ $f(x) ; Q= f (x)$

Integrating factors I.F = $e^{{\intop p dx}}$

I.F = $e^{{\intop \frac{-2}{x}dx}}$ = $e^{{-2\ln x}}$ = $e^{{\ln x (\frac{1}{x^2}) }}$ = $\frac{1}{x^2}$

$\frac{1}{x^2}$ $\frac{dv}{dx}$ $\frac{-2}{x^3}.v= \frac{-2}{x^2}$

= $\frac{dv}{x^2}- \frac{2dx}{x^3}.v =\frac{dx}{x^2}$

$\intop d(\frac{v}{x^2}) = \intop \frac{dv}{x^2} = \frac{v}{x^2} = x ^{{-1}} + c$

$\frac{1}{y^2}. \frac{1}{x^2} = \frac{-1}{x} + c$

$\frac{1}{x^2y^2} + \frac{1}{x} = c$