Solutions of homogeneous linear differential equation form a vector space 
Show that the solutions of a homogeneous linear differential equation $y''+a(x)y'+b(x)y = 0$ form a vector space. What is its dimension?

I understand that the dimension is 2 and that 0 is a solution to the differential equation ($0''+a(x)*0'+b(x)*0=0$).
How does one go about proving the other two properties of a vector space: closed under addition and closed under multiplication?
 A: Let $y_1,y_2$ be solutions for the equation
$$y''+ay'+by=0.$$
Then
$$y_1''+ay_1'+by_1=0$$
$$y_2''+ay_2'+by_2=0$$
Add the first and the second equation we have
$$(y_1+y_2)''+a(y_1+y_2)'+b(y_1+y_2)=0.$$
Then $y_1+y_2$ is solution too.
Let $\lambda$ be a real number
$$\lambda y_1''+a\lambda y_1'+b\lambda y_1=\lambda(y_1''+ay_1'+by_1)=0.$$
Then $\lambda y_1$ is solution too.
A: Let $V = \{y \mid y'' + ay' + by = 0\}$.
To show that $V$ is closed under scalar multiplication, let $k \in \mathbb{R}$ and $y(x) \in V$. You want to show that $ky \in V$, so you need to verify that $ky$ satisfies the defining property of $V$; namely, you need to show that $(ky)'' + a(ky)' + b(ky) = 0$ by using that fact that $y \in V$, so $y'' + ay' + by = 0$.
Likewise, to show that $V$ is closed under vector addition, let $y_1, y_2 \in V$. You want to show that $y_1 + y_2 \in V$, so you need to verify that $y_1 + y_2$ satisfies the defining property of $V$; namely, you need to show that $(y_1 + y_2)'' + a(y_1+y_2)'+b(y_1+y_1) = 0$ by using the fact that $y_1, y_2 \in V$, so $y_1'' + ay_1' + by_1 = 0$ and $y_2'' + ay_2' + by_2 = 0$.
