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How would you go about solving the following system of ODEs:

\begin{align*} & x''(t) - \frac{2}{y}x'(t) \ y'(t) = 0 \ & y''(t) + \frac{1}{y} \big(x'(t) - y'(t)\big) = 0 \end{align*}

Any help would be very much appreciated!

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I have managed to reduce it to the single equation $y''(t) + \frac{1}{y}((y'(t))^2 + 1) = 0.$ – Alex Kite Apr 7 '12 at 2:42
So the first equation gives $\ln x' = 2\ln y+C$, plugging that into the second equation gives $y''(t)+\frac{1}{y}(C'y'(t)^2-y')$. How did you get the $+1$ term? – Alex R. Apr 7 '12 at 2:48
up vote 0 down vote accepted

From the first equation we can conclude :

$\ln x'(t)= 2\ln y +C$ , so $x'(t)=C_1\cdot y^2$

plugging this into second equation gives :

$y''(t)+\frac{1}{y}(C_1 \cdot y^2-y'(t))=0$

Now substitute $y'(t)=v$ , where $v$ is a function in terms of variable $y$ ,so:

$y''(t)=v'_{y}\cdot v$

Hence :

$v'_{y}\cdot v+\frac{1}{y}(C_1 \cdot y^2-v)=0$

this equation is equivalent to the first order non-linear ODE :

$v'_y+C_1\cdot y \cdot v^{-1} -\frac{1}{y}=0$

which can be solved using numerical methods .

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