# Converting second order ODE into first order ODE

I am given the differential equation $y'' + y/|y|^3 = 0$ with $y(0) = (1,0)$ and $y'(0) = (0,0.25)$. I need to convert this second order ODE into a system of first order ODEs.

So I set $y_1 = y$ and $y_2 = y' \implies y_{1}^{'} = y_2$. Then, $y_{2}^{'} = y^{''} = -y/|y|^3 = -y_1/|y_1|^3$. Since $y_1(0) = (1,0) \text{and} y_2(0) = (0,0.25)$ but then the vectors trip me up in getting the final system $(y_{1}^{'} = y_2, y_2 = " ")$. How do I deal with the vector here?

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Sorry, I misread your question. Hold on, I'll explain. – bonext Oct 29 '12 at 16:13

Basically, in the very beginning you have $y=(y^1,y^2)$ (notice I put indices on top!), so when you do $y'$ it is actually $y'=((y^1)',(y^2)')$.
After that, you say $y_1=y$ and $y_2=y_1'$, so what you really get is: $y_1=(y^1,y^2)=(y_1^1,y_1^2)$ and $y_2=((y^1)',(y^2)')=(y_2^1,y_2^2)$ and hence $y_2^1=(y_1^1)'$, $y_2^2=(y_1^2)'$.
The total vector form of your now "normalized" system will be expressed in terms of the vector $\mathbf{y}=(y_1,y_2)=(y_1^1,y_1^2,y_2^1,y_2^2)$, so the I.C. will be just $\mathbf{y}(0)=(y_1^1(0),y_1^2(0),y_2^1(0),y_2^2(0))=(y^1(0),y^2(0),(y^1)'(0),(y^2)'(0))=(y(0),y'(0))=(1,0,0,0.25)$
You have to solve for the whole $\mathbf{y}=(y_1^1, y_1^2, y_2^1, y_2^2)$ as long as you actually have a Cauchy I.V. problem of the form $\mathbf{y'}=f(\mathbf{y}),\quad \mathbf{y}(0)=\mathbf{y_0}$ – bonext Oct 30 '12 at 7:53