Solution
= Solution
An <integrating factor> is $e^{3x}$. Multiplication gives $(e^{3x}y)'=8e^{3x}$, so $y=8/3+Ce^{-3x}$. The <initial condition> gives $C=4/3$. Therefore
$$
\boxed{y(x)=\frac83+\frac43e^{-3x}.}
$$
= Solution
An <integrating factor> is $e^{3x}$. Multiplication gives $(e^{3x}y)'=8e^{3x}$, so $y=8/3+Ce^{-3x}$. The <initial condition> gives $C=4/3$. Therefore
$$
\boxed{y(x)=\frac83+\frac43e^{-3x}.}
$$