Formulate two equations as a system of first-order ODEs I have two equations that I want to formulate into a system of first-order ODEs.
\begin{cases}
  \displaystyle
  x''(t) = \frac{x(t)}{\Big( \sqrt{x(t)^2 + y(t)^2} \Big)^{3}} \\
  \displaystyle
  y''(t) = \frac{y(t)}{\Big( \sqrt{x(t)^2 + y(t)^2} \Big)^{3}} 
\end{cases}
and I also know the following information:
$x(0)=-1$
$x'(0)=0$
$y(0) = 0$
$y'(0) = -1$




I have tried to solve this math problem by introducing $z$ and $w$ so that: 
\begin{cases} 
z(t) = x'(t)\\
w(t) = y'(t)\\
\displaystyle
z'(t) = x''(t) = \frac{x(t)}{(\sqrt{x(t)^{2} + y(t)^{2}})^{3}} \\
\displaystyle
w'(t) = y''(t) = \frac{y(t)}{(\sqrt{x(t)^{2} + y(t)^{2}})^{3}} \\
\end{cases}
and then updating the initial condition to accordingly:
$z(0) = x'(0) = 0$
$z'(0)= x(0)/(\sqrt{x(0)^{2} + y(0)^{2} })^{3} = -1$
$w(0) = y'(0) = -1$
$w'(0) = y(0)/(\sqrt{x(0)^{2} + y(0)^{2} })^{3} = 0$ 
Is this a valid solution to the problem?
The math book I use does not show what the correct answer is suppose to be.

Yes, your work is correct. 
Usually the system is written as 
$$\begin{cases} 
x'(t)=z(t) \\
y'(t)=w(t)  \\
\displaystyle
z'(t) = \frac{x(t)}{(\sqrt{x(t)^{2} + y(t)^{2}})^{3}} \\
\displaystyle
w'(t) = \frac{y(t)}{(\sqrt{x(t)^{2} + y(t)^{2}})^{3}} \\
\end{cases}$$
Or they use $u_1, u_2, u_3, u_4$ for $x,y,z,w$
$$\begin{cases} 
u_1'=u_3 \\
u_2'=u_4  \\
\displaystyle
u_3' = \frac{u_1}{(\sqrt{u_1^{2} + u_2^{2}})^{3}} \\
\displaystyle
u_4' = \frac{u_2}{(\sqrt{u_1^{2} + u_2^{2}})^{3}} \\
\end{cases}$$
