Answer
$y=\frac{{(3x+2)}^4}{12}-\frac{1}{3}$
Work Step by Step
$\frac{dy}{dx}=(3x+2)^3;$ $y(0)=1$
$dy=(3x+2)^3dx$
$\int dy=\int(3x+2)^3dx$
$y+C=\frac{1}{3}\int3(3x+2)^3dx$
$y=\frac{1}{3}(\frac{{(3x+2)}^4}{4})+C$
Substitute in initial condition to solve for $C$
$1=\frac{4}{3}+C$
$C=-\frac{1}{3}$
$y=\frac{{(3x+2)}^4}{12}-\frac{1}{3}$
