Hello,

I have the DE:

$\displaystyle y''(t) + y = 0$

with IC's:

$\displaystyle y'(0) = 1$hs

$\displaystyle y(0) = 0$

If I convert this to a system of 1st order ODES is it just:

$\displaystyle \frac{dy_1}{dt} = y_2, \ \ \ \frac{dy_2}{dt} = -y_1 $

How about if the DE is:

$\displaystyle y''(t) + y = \sin(2t)$ and the same IC's.

Thanks