# Any idea on how to solve this system of coupled ODEs?

I'm trying to find solutions for the system of ODEs

$$y_1'(t) = y_1(t)y_2(t) \\ y_2'(t) = 2y_2(t)^2 - y_1(t)^6$$

And I'm assuming $y_1(t), y_2(t) > 0$. This comes from trying to find the characteristic curves of the vector field $<xy,2y^2 - x^6>$ defined over $(\mathbb{R}^+)^2$. Of course this can be easily decoupled into a 2nd order ODE in $y_1(t)$ which I however find no way to deal with. Can you suggest possible approaches?

-

$y_2(t)=\dfrac{y_1'(t)}{y_1(t)}$

$y_2'(t)=\dfrac{y_1''(t)}{y_1(t)}-\dfrac{y_1'(t)^2}{y_1(t)^2}$

$\therefore\dfrac{y_1''(t)}{y_1(t)}-\dfrac{y_1'(t)^2}{y_1(t)^2}=\dfrac{2y_1'(t)^2}{y_1(t)^2}-y_1(t)^6$

$\dfrac{y_1''(t)}{y_1(t)}-\dfrac{3y_1'(t)^2}{y_1(t)^2}=-y_1(t)^6$

$y_1(t)y_1''(t)-3y_1'(t)^2=-y_1(t)^8$

$y_1\dfrac{d^2y_1}{dt^2}-3\left(\dfrac{dy_1}{dt}\right)^2=-y_1^8$

Let $\dfrac{dy_1}{dt}=u$ ,

Then $\dfrac{d^2y_1}{dt^2}=\dfrac{du}{dt}=\dfrac{du}{dy_1}\dfrac{dy_1}{dt}=u\dfrac{du}{dy_1}$

$\therefore y_1u\dfrac{du}{dy_1}-3u^2=-y_1^8$

$\dfrac{du}{dy_1}-\dfrac{3u}{y_1}=-\dfrac{y_1^7}{u}$

Let $v=u^2$ ,

Then $\dfrac{dv}{dy_1}=2u\dfrac{du}{dy_1}$

$\therefore\dfrac{1}{2u}\dfrac{dv}{dy_1}-\dfrac{3u}{y_1}=-\dfrac{y_1^7}{u}$

$\dfrac{dv}{dy_1}-\dfrac{6u^2}{y_1}=-2y_1^7$

$\dfrac{dv}{dy_1}-\dfrac{6v}{y_1}=-2y_1^7$

I.F. $=e^{\int-\frac{6}{y_1}dy_1}=e^{-6\ln y_1}=\dfrac{1}{y_1^6}$

$\therefore\dfrac{d}{dy_1}\left(\dfrac{v}{y_1^6}\right)=-2y_1$

$\dfrac{v}{y_1^6}=\int-2y_1~dy_1$

$\dfrac{u^2}{y_1^6}=C_1^2-y_1^2$

$\dfrac{u}{y_1^3}=\pm\sqrt{C_1^2-y_1^2}$

$\dfrac{dy_1}{dt}=\pm y_1^3\sqrt{C_1^2-y_1^2}$

$\pm\dfrac{dy_1}{y_1^3\sqrt{C_1^2-y_1^2}}=dt$

$\int\pm\dfrac{dy_1}{y_1^3\sqrt{C_1^2-y_1^2}}=\int dt$

$\pm\dfrac{\sqrt{C_1^2-y_1^2}}{2C_1^2y_1^2}\pm\dfrac{1}{2C_1^3}\ln\dfrac{C_1+\sqrt{C_1^2-y_1^2}}{y_1}=t+C_2$

Bring the following back to youself to think.

-

Here is one possible solution,

$$y_1(t) =0\,, \,y_2(t) = \frac{1}{ C - 2\,t }\,.$$

-
This fails the condition that $y_1(t)\gt0$. – Did Oct 14 '12 at 9:34