Finding eigenvalues and corresponding eigenspaces of a polynomial linear operator I have the following question I am struggling with.

Let $\mathbb{R}[x]$ be the real vector space of polynomials in the variable $x$ with real coefficients and define a linear map $\phi:\mathbb{R}[x]\to\mathbb{R}[x]$ by $$\phi(p):=x^2 \frac{d^2p}{dx^2} + x \frac{dp}{dx}.$$
1.Find all eigenvalues of $\phi$ and the corresponding eigenspaces
2.Is the map $\phi$ injective?
3.Is the map $\phi$ surjective?
4.Find the most general polynomial $p\in \mathbb{R}[x]$ satisfying $$\phi(p)=x+3x^3.$$

This is what I have so far:
Let $p(x)=\sum_{n=0}^{\infty} a_nx^n\in \mathbb{R}[x]$, then $$\phi(p)(x)=\sum_{n=2}^{\infty}n(n-1)a_nx^n + \sum_{n=1}^{\infty} na_nx^n \\ =\sum_{n=1}^{\infty} n^2a_nx^n$$
I am trying to equate this to $\lambda p(x)$ but I am stuck with this: $$\sum_{n=1}^{\infty} n^2a_nx^n=\sum_{n=0}^{\infty} \lambda a_nx^n.$$

Firstly, the polynomials have finite degree, so it should be $p(x)=\sum_{k=0}^{\color{red}n}a_kx^k$. 
For eigenvalues
\begin{align*}
\phi(p)&=\lambda p\\
x^2p^{''}+xp^{'}&=\lambda p\\
\sum_{k=2}^n{k(k-1)a_k}x^k+\sum_{k=1}^nka_kx^k&=\lambda\sum_{k=0}^na_kx^k\\a_1x+x^2(4a_2)+x^3(9a_3)+\dotsb+x^{k}(k^2a_k)+\dotsb&=\lambda(a_0+a_1x+a_2x^2+\dotsb+a_kx^k+\dotsb)
\end{align*}geq 1
By comparing the coefficients of $x^k$, we get the following system:
\begin{align*}
\lambda a_0&=0\\
\lambda a_1&=a_1\\
\lambda a_2&=2^2a_2\\
\vdots&=\vdots\\
\lambda a_k&=k^2a_k\\
\vdots&=\vdots
\end{align*}
If $a_0 \neq 0$,then $\lambda=0$. In which case, for all $i \geq 1$,we have $a_i=0$. This tells us that $\color{red}{p(x)=a_0}$ is an eigenvector with eigenvalue $\color{red}{\lambda=0}$.
Second  case is when $a_0=0$. Then from the second equation we get that either $\lambda=1$ or $a_1=0$. If $\lambda=1$, then $a_i=0$ for all $i \geq 2$. As before, $\color{red}{p(x)=0+x=x}$ is an eigenvector with eigenvalue $\color{red}{\lambda=1}$.
Now check the case when $a_1=0$ and see what happens. Hopefully you can take it from here. 
You should see a pattern that it is enough to check what happens when $p(x)=x^n$. Observe that
$$\phi(x^n)=n^2x^n.$$
Thus $\color{red}{p(x)=x^n}$ is an eigenvector with eigenvalue $\color{red}{\lambda=n^2}$.
