Answer
See below
Work Step by Step
Let $y_1,y_2$ be the solutions of $y'+3y=0$
then $y_1'+3y_1=0\\
y_2'+3y_2=0$
$\Rightarrow (y_1+y_2)'+3(y_1+y_2)\\
=y_1'+y_2'+3y_1+3y_2\\
=(y_1'+3y_1)+(y_2'+3y_2)\\
=0+0\\
=0$
If we assume that $y_1+y_2$ is also the solutions to the given equation, then $y_1+y_2 \in S$.
$\Rightarrow y_1+y_2$ is 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.
$\Rightarrow \lambda (y_1)'+3(\lambda y_1)\\
=\lambda y_1'+3\lambda y_1\\
=\lambda(y_1'+3y_1)\\
=\lambda .0\\
=0$
Assume $\lambda y_1$ is a solution to the given equation, then $ky_1 \in S$.
Therefore $S$ is closed under addition.
