Answer to Question #216104 in Differential Equations for Karylle

Question

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

Answer 1

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

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

Let y = vx, then $\frac{dy}{dx } = v+x\frac{dv}{dx}$

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

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

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

Integrating both sides,

$\int \frac{3vdv}{1-2v^3-3v^2} =\int \frac{dx}{x}$

Doing partial fractions,

$\frac{3v}{1-2v^3-3v^2} = \frac{1}{9(v+1)} -\frac{1}{3(v+1)^2}-\frac{2}{9(2v-1)}$

Now integrate both sides,

$\int \frac{1}{9(v+1)} -\frac{1}{3(v+1)^2}-\frac{2}{9(2v-1)} dv =\int\frac{dx}{x}$

$\frac{1}{3}ln|v+1| +\frac{1}{v+1} -\frac{1}{3}ln|2v-1| = ln|x| + ln|C|$

$\frac{1}{3}ln|\frac{v+1}{2v-1}| +\frac{1}{v+1} = ln|Cx|$

Simply putting $v = \frac{y}{x}$

$\frac{1}{3}ln|\frac{y+x}{2y-x}| +\frac{x}{y+x} = ln|Cx|$