# What is a solution to the differential equation y'=-e^(3x)?

$y = - \frac{1}{3} {e}^{3} x + C$
If $y ' = - {e}^{3} x$ then integrating gives;
$y = \int \left(- {e}^{3} x\right) \mathrm{dx}$
$\therefore y = - \frac{1}{3} {e}^{3} x + C$