## Calculus (3rd Edition)

$y=-t^2+3t-5$
$\frac{dy}{dt}=3-2t;$ $y(0)=-5$ $dy=(3-2t)dt$ $\int dy=\int(3-2t)dt$ $y=3t-t^2+C$ Substitute in initial condition $-5=3(0)-0^2+C$ $C=-5$ $y=-t^2+3t-5$