Checking if the set of solutions to $y''+ 4y'+ 8y = 0$ is a vector space with the usual operations. I've read a question that ask me to check that the set of functions on the line that has second derivative and verify the equation $y''+ 4y'+ 8y = 0$ is a vector space with the usual operations.
I know that it's easy to prove that is a "subspace". But to prove that is a space, for example, I tried:
Being $y$ where $y''+ 4y'+ 8y = 0 \Rightarrow  y$ belongs to "set of functions ...." 
Being $x$ where $x''+ 4x'+ 8x = 0 \Rightarrow  x$ belongs to "set of functions ...."  
Being $z$ where $z''+ 4z'+ 8z = 0 \Rightarrow  z$ belongs to "set of functions ...." 
I just know that
$(y+x)' + 4(y+x)'+ 8(x+y) =0$ belongs to "set of functions ...." 
the neutral, vector $0$, can be any $z''+ 4z'+ 8z = 0?$
opposite  $(y-y)' + 4(y-y)'+ 8(y-y)= 0-0=0$
How can I wwork with this in associativity, commutativity, the opposite existence ..
A1) $(u+v)+w = u+(v+w)$
A2) $u+v = v+u$
 A: It is easier to understand this type of problem in the context of operator theory. Our underlying space, or "set of functions," is $C^\infty(\mathbb R)$, the set of all infinitely-differentiable real-valued functions defined on $\mathbb R$. For example, if $y(x)=e^x$, then $y = \in C^\infty(\mathbb R)$ (this is clear, as $y'=y$).
It's straightforward to show that $C^\infty(\mathbb R)$ forms a vector space over the field of real numbers under the operation of function addition (i.e. $(f+g)(x) = f(x) + g(x)$), with the zero function as the identity and the additive inverse of a function $f$ being $-f$ where $(-f)(x) = -f(x)$. The axioms of associativity, commutativity, etc. follow from those of addition of real numbers.
Define an operator $L:C^\infty(\mathbb R)\to C^\infty(\mathbb R)$ by
$$Ly = y'' + 4y' + 8y. $$
For example, if $y(x) = x^2$, then $y'(x) = 2x$ and $y''(x)=2$, so
$$Ly(x) = (y'' + 4y' + 8y)(x), $$
and by linearity of differentiation,
$$(y'' + 4y' + 8y)(x) = y''(x) + 4y'(x) + 8y(x) = 2 + 4(2x) + 8(x^2) = 8x^2 + 8x + 2. $$
Let $$S = \{y\in C^\infty(\mathbb R) : Ly = 0\}.$$
Since $S$ is a subset of $C^\infty(\mathbb R)$, in order to show that $S$ is a subspace, we need only show that it contains the identity (which is clear, since the derivative of a constant function is zero, and so $L0 = 0$), and that it is closed under addition and scalar multiplication. For the former, if $y_1,y_2\in S$ then $Ly_1=Ly_2=0$, so
$$
\begin{align*}
L(y_1+y_2) &= (y_1+y_2)'' + 4(y_1+y_2)' + 8(y_1+y_2)\\
&= y_1'' + y_2'' + 4y_1' + 4y_2' + 8y_1 + 8y_2\\
&= (y_1'' + 4y_1' + 8y_1) + (y_2'' + 4y_2' + 8y_2)\\
&= Ly_1 + Ly_2,\\
&= 0
\end{align*}
$$
so that $y_1+y_2\in S$. For the latter, observe that if $c\in\mathbb R$ then
$$L(cy_1) = (cy_1)'' + 4(cy_1)' + 8(cy_1) = c(y_1'' + 4y_1' + 8y_1) = cLy_1 = 0,$$
so that $cy_1\in S$.
