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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?

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up vote 2 down vote accepted







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

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

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


Let $v=u^2$ ,

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




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





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


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


Bring the following back to youself to think.

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Here is one possible solution,

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

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

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