Solve the homogeneous ode 2xy²dy-(x³+2y³dx=0?

May 10, 2018

$\frac{2 {y}^{3}}{3 {x}^{3}} = \ln \left(c x\right)$ore^((2y^3)/(3x^3) =cx

Explanation:

Here,

2xy²dy-color(red)((x³+2y³)) dx=0

=>2xy²dy=(x³+2y³)dx

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{3} + 2 {y}^{3}}{2 x {y}^{2}}$

=>(dy)/(dx)=(x^3(1+2(y/x)^3))/(x^3(2(y/x)^2)

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 + 2 {\left(\frac{y}{x}\right)}^{3}}{2 {\left(\frac{y}{x}\right)}^{2}}$

Let us subst.

color(blue)(y/x=v=>y=vx=>(dy)/(dx)=v+x(dv)/(dx)

So,

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

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

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

$\implies 2 {v}^{2} \mathrm{dv} = \frac{1}{x} \mathrm{dx}$

Integrating both sides :

$\implies 2 \int {v}^{2} \mathrm{dv} = \int \frac{1}{x} \mathrm{dx} + c '$

$\implies 2 \cdot {v}^{3} / 3 = \ln x + \ln c \ldots \to \left[c ' = \ln c\right]$

$\implies \frac{2}{3} {v}^{3} = \ln \left(c x\right)$

Subst. back, color(blue)(v=y/x,we get

$\frac{2}{3} \left({y}^{3} / {x}^{3}\right) = \ln \left(c x\right)$

$\implies \frac{2 {y}^{3}}{3 {x}^{3}} = \ln \left(c x\right)$

=>e^((2y^3)/(3x^3) =cx