Answer
See below
Work Step by Step
Let $y_1,y_2$ be the solutions of $y'+3y=6x^3+5$
then $y_1'+3y_1=6x^3+5\\
y_2'+3y_2=6x^3+5$
$\Rightarrow (y_1+y_2)'+3(y_1+y_2)=6x^3+5\\
y_1'+y_2'+3y_1+3y_2=6x^3+5\\
(y_1'+3y_1)+(y_2'+3y_2)=6x^3+5\\
6x^3+5+6x^3+5=6x^3+5\\
6x^3+5=0$
If we assume that $y_1+y_2$ is also the solutions to the given equation, then it is not true.
$\Rightarrow u+v$ is not closed under addition.
Set of Rational numbers is closed under usual addition.
Because our scalars can come from set of real numbers
In particular, let $\lambda$ be a scalar.
$\Rightarrow \lambda (y_1)'+3(\lambda y_1)=6x^3+5\\
\lambda y_1'+3\lambda y_1=6x^3+5\\
\lambda (y_1'+3y_1)=6x^3+5\\
k(6x^3+5)=6x^3+5$
Assume $\lambda y_1$ is a solution to the given equation, then it is generally not true.
Therefore $S$ is not closed under addition.