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

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

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

\(\displaystyle Solution\)

\(\displaystyle y=a\ sin\ bx\)

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

\(\displaystyle What\ is\ the\ rule\ for\ finding\ the\ derivative?\)

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

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

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

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

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