## Calculus 10th Edition

$y$ = $3t^{3} + C$
$\frac{dy}{dt}$ = $9t^{2}$ To get original equation, we have to integrate the aforementioned differential equation. $y=\int dy=\int9t^{2} dt$ = $\frac{9t^{2+1}}{3} + C$ = $3t^{3} + C$ Hence, $y$ = $3t^{3} + C$