# Thread: Need help on a system of ODEs?

1. ## Need help on a system of ODEs?

dx/dt = x(x^3 - 2y^3)

dy/dt = y(2x^3 - y^3)

2. $\frac{dy}{dx} = \frac{y(2x^3-y^3)}{x(x^3-2y^3)}$. Notice that both the numerator and denominator are of degree 4,ie homogeneous. Do you know how to solve a homogeneous DE?

Hint: take $y=vx$

3. Originally Posted by Sambit
$\frac{dy}{dx} = \frac{y(2x^3-y^3)}{x(x^3-2y^3)}$. Notice that both the numerator and denominator are of degree 4,ie homogeneous. Do you know how to solve a homogeneous DE?

Hint: take $y=vx$
after i sub y=tx
i got dy/dx={t(2-t^3)}/{1-2t^3}

but then what should be doing next? thanks alot

4. If y = xv then your ODE becomes

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

which separate to

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

or

$\left(\dfrac{1}{v} - \dfrac{3v^2}{v^3+1} \right) dv = \dfrac{dx}{x}$ (this easily integrates!)

5. what is the 'v' means and what is the reason to sub 'y=vx'?

it was dy/dx if sub y=vx
dy= v dx ?

thanks

6. If you have an ODE of the form

$\dfrac{dy}{dx} = F \left(\dfrac{y}{x}\right)$

(called homogeneous) then a substitution $y = x v$ will give another ODE that separates.

Note $y = y(x) \; \text{and}\; v = v(x)$

7. oh I see!! Thank you so much for your clear explanation.

8. y=y(x)
y=vx v=y/x 1/v=x/y

dy/dx = x(dv/dx)

x(dv/dx)={ v(2-v^3)}/ (1-2v^3)

is this form correct? thx

9. Product rule

$\dfrac{dy}{dx} = x \dfrac{dv}{dx} + v$.

10. Thank you so much for your remind!
after integration, I got

v/(v^3 +1) = x + e^c
v/(v^3 +1) = x + A ,where A is constant