Computing the kernel of a linear operator defined on a space of polynomials 
I am trying to find the Kernel for this vector space, but from what I understand the elements of my kernel must be polynomials since it is a subspace of my vector space. The usual answer I can find is in terms of exp. function but I cant find such polynomial.
 A: The solution of the differential equation
$$3p'''+2p''=0$$
is
$$p=Ae^{-2t/3}+Bt+C\ .$$
This is a polynomial (of degree at most $4$) if and only if $A=0$.  So the kernel is
$$\{Bt+C\mid B,C\in{\Bbb F}\}\ .$$
Alternative solution.  Take a general polynomial
$$p=at^4+bt^3+ct^2+dt+e\ ,$$
substitute into the differential equation, and see what this tells you about $a,b,c,d,e$.
A: Since this is about polynomials, you should ignore exponential functions; they are not considered in this problem. You can write your linear operator in a basis of this vector space, and find its kernel easily.
The easiest basis to use is $[1,t,t^2,t^3,t^4]$. With respect to that basis $f$ has the matrix
$$
  A=\begin{pmatrix}0&0&2&3&0\\0&0&0&2&3\\0&0&0&0&2\\0&0&0&0&0\\0&0&0&0&0
  \end{pmatrix}.
$$
The equation $A\cdot x=0$ is now easy to solve. If $2$ is invertible one easily gets $0=x_5=x_4=x_3$, and $x_1,x_2$ can be freely chosen. Since $x=(x_1,x_2,0,0,0)$ represents (in the given basis) the polynomial $x_0+x_1t$, one gets as $2$-dimensional kernel the subspace $F[t]_1$ of polynomials of degree at most$~1$. 
However, since the field was not specified and formal differentiation was mentioned, I think you should allow for the possibility that $F$ has characteristic$~2$; in that case the solution has $3$ free parameters rather than $2$, and the kernel has dimension$~3$ (it is $F[t]_2$).
