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 $x,y,z,w$
$$\begin{cases} u_1'=u_3 \\ u_2'=u_4 \\ \displaystyle u_3' = \frac{u_1}{(\sqrt{u_1^{2} + u_2^{2}})^{3}} \\ \displaystyle u_4' = \frac{u_2}{(\sqrt{u_1^{2} + u_2^{2}})^{3}} \\ \end{cases}$$