Answer
$$y = - \frac{4}{3}{e^{ - 3x}} + C$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = 4{e^{ - 3x}} \cr
& {\text{Separating variables leads to}} \cr
& dy = 4{e^{ - 3x}}dx \cr
& {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr
& \int {dy} = \int {4{e^{ - 3x}}} dx \cr
& \int {dy} = 4\int {{e^{ - 3x}}} dx \cr
& {\text{rewrite the integrand on the right side}} \cr
& \int {dy} = - \frac{4}{3}\int {{e^{ - 3x}}} \left( { - 3} \right)dx \cr
& {\text{integrate using }}\int {{e^u}} du = {e^u} + C.{\text{ let }}u = - 3.{\text{ then}} \cr
& y = - \frac{4}{3}{e^{ - 3x}} + C \cr
& \cr
& {\text{verifying that the solution satisfies the original differential equation}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ { - \frac{4}{3}{e^{ - 3x}} + C} \right] \cr
& \frac{{dy}}{{dx}} = - \frac{4}{3}\left( { - 3{e^{ - 3x}}} \right) + 0 \cr
& \frac{{dy}}{{dx}} = 4{e^{ - 3x}} \cr} $$
