Trig derivative questions.

$\displaystyle I\ have\ an\ equation\ and\ solution.\ I'm\ unclear\ on\ a\ couple\ of\ things.$

$\displaystyle An\ analysis\ of\ the\ motion\ of\ a\ particle\ resulted\ in\ the\ differential\ equation$

$\displaystyle \frac{d^2y}{dt^2}\ +\ b^2y\ =\ 0$

$\displaystyle Show\ that\ y=a\ sin\ bx\ is\ a\ solution\ of\ this\ equation.$

$\displaystyle Solution$

$\displaystyle y=a\ sin\ bx$

$\displaystyle \frac{dy}{dt}\ =\ ab\ cos\ bt\ Why\ has\ x\ become\ t\ and\ why\ is\ there\ another\ b\ in\ the\ equation?$

$\displaystyle What\ is\ the\ rule\ for\ finding\ the\ derivative?$

$\displaystyle \frac{d^2y}{dt^2}\ =\ -ab^2\ sin\ bt$

$\displaystyle \frac{d^2y}{dt^2}\ =\ -b^2(a\ sin\ bt)$

$\displaystyle \frac{d^2y}{dt^2}\ =\ -b^2y $

$\displaystyle \frac{d^2y}{dt^2}+b^2y=-b^2y+b^2y=0$

$\displaystyle y=a\ sin\ bt\ is\ the\ solution\ of \ the\ given\ differential\ equation.$