Do you understand what a "vector space" is?
In order to show that a set, with a given "addition" and "scalar multiplication" is a vector space you must show three things:
1) It is non empty. You can do that by showing that it contains the "zero vector" (the additive identity). Here, since the addition is ordinary addition of functions, the additive identity is the 0 function: f(x)= 0 for all x. Show that y= 0 satisfies a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= 0 for any functions. (The superscript in parentheses indicates the derivative.)
2) It is "closed under scalar multiplication". To do that show that if y(t) satisfies a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= 0 then so does ay(t) for any number a. That's just a matter of factoring out a.
3) It is "closed under addition". If y= u(t) and y= v(t) both satisfy a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= 0 then so does y= u+ v. That can be done because the nth derivative is itself "linear"- the nth deriviative of u+ v is the nth derivative of u plus the nth derivative of v.
To show that the solution set of a linear homogeneous nth order equation is a vector space of dimension n, use the "existence and uniqueness theorem" to define the n functions, u_1 that satisfies the given equation with initial conditions u_1(0)= 1, u_1'(0)= u_1''(0)= \cdot\cdot\cdot= u_1^{(n-1)}(0)= 0, u_2(0)= 0, u_2'(0)= 1, u_2''(0)= \cdot\cdot\cdot= u_2^{(n-1)}(0)= 0, etc., with "u_i(t)" being the solution to the differential equation having its ith derivative equal to 1 at t= 0 and all other values 0.
Those are sometimes referred to as the "fundamental solutions" to the differential equation. They have this very nice property: if y(t) is any solution to the differential equation, let A_1= y(0), A_2= y'(0), ... , A_{n}= y^{(n-1)}(0). Then it is easy to show that y(t)= A_1u_1(t)+ A_2 u_2(t)+ \cdot\cdot\cdot+ A_{n}u_{n-1}(t). That is, any solution to the differential equation can be written as a linear combination of the "fundamental solutions" so they form a basis for the vector space of all solutions showing that the vector space has dimension n. Any solution can be written as a linear combination of those n independent solutions.
Now, for the non-homogeneous equation, let Y(t) be any function satisfying the equation a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= f(x). Then it is easy to show that if y is any other function satisfying that equation, u(t)= y(t)-Y(t) satisfies the equation a_n(t)y^{(n)}+ a_{n-1}y^{(n-1)}+ \cdot\cdot\cdot+ a_1y'+ a_0y= 0. That is, that y(t)= u(t)+ Y(t), it is the sum of a function in the vector space of solutions to the homogeneous equation plus that one given solution to the entire equation.
