Question

Bernoulli's DE

(2y^3-x^3)dx+3x^2ydy=0

Answer 1

Solution;

Rewrtie the equation as follows:

$\frac{dy}{dx}$ +P(x)y=Q(x)yⁿ

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

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

The equation is not a Bernoulli's so we solve as an homogenous equation.

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

Take

y=Vx

$\frac{dy}{dx}$ =v+x$\frac{dv}{dx}$

v+x$\frac{dv}{dx}$ =-$\frac{2v^3x^3-x3}{3vx^3}$

Divide the R.H.S with x³

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

Separate the variables;

$\frac{3v}{1-2v^3-3v^2}$dv=$\frac1x$ dx

Integrating both sides;

3($-\frac{ln(2v-1)}{9}$ +$\frac{ln(v+1)}{9}$+$\frac{3}{3v+3}$ =ln(x)+C

But v=$\frac yx$

-$\frac{ln(\frac yx-1)}{3}$ +$\frac{ln(\frac yx+1)}{3}$+$\frac{3}{3\frac yx+3}$ =ln(x)+C