Prove solutions to 3rd order differential equation form 3-dimensional vector space Prove that the solutions to $ \dfrac{ \mathrm{d^{3}y}}{ \mathrm{dx^{3}}} -6 \dfrac{ \mathrm{d^{2}y}}{ \mathrm{dx^{2}}}+11 \dfrac{ \mathrm{dy}}{ \mathrm{dx}}-6y=0$ form a 3-dimensional vector space:

Its simple to verify the set of solutions, $V$, form a vector space by noting $ \dfrac{ \mathrm{d}}{ \mathrm{dx}} ( \alpha f + \beta g) = \alpha \dfrac{ \mathrm{df}}{ \mathrm{dx}} + \beta \dfrac{ \mathrm{dg}}{ \mathrm{dx}} $, however I am unsure how to show $\dim V= 3$. 

I know that in general, an nth order linear differential equation has n linearly independent solutions, but that has not been used in the previous part of the book this question is from and seems lengthy to prove, so is there an obvious way?

(Its clear $e^x$ is a solution but not sure how this helps).
Hint: The characteristic polynomial has three solutions $\lambda=1,2,3$:

Hence the general solution is $y=c_1\exp(x)+c_2\exp(2x)+c_3\exp(3x)$. The Wronskian determinant is non-zero, hence these are three linearly independent solutions spanning a three dimensional vector space (show by checking definition of vector space). This can be directly guessed by looking at the integer factors of $6$.
Hint: The characteristic polynomial has three solutions $\lambda=1,2,3$:

Hence the general solution is $y=c_1\exp(x)+c_2\exp(2x)+c_3\exp(3x)$. The Wronskian determinant is non-zero, hence these are three linearly independent solutions spanning a three dimensional vector space (show by checking definition of vector space). This can be directly guessed by looking at the integer factors of $6$.