"Differential Operator" Over Polynomial Space

Question

Let's suppose we are given the differential operator $T \colon \mathcal{P}_2(\mathbb{C}) \longrightarrow \mathcal{P}_3(\mathbb{C})$, over the space of quadratic polynomials with complex coefficients, such that $T(p(t)) := p(t) + t^2\,p'(t)$, and we are asked to find it's kernel. Of course, setting $p(t) := a_0 + a_1\,t + a_2\,t^2$, where $a_0,\,a_1,\,a_2 \in \mathbb{C}$, one can easily find that $\text{Ker}(T) = \left\{0\right\}$, the zero polynomial, with the polynomial equality.
But, can we actualy solve the equation $p(t) + t^2\,p'(t) = 0$? I know we're going to find an exponential solution of the form $k\exp(1/t)$, but can we take $k = 0$ (and hence $p(t) = 0$) to solve this problem?
Thanks in advance!

Answer 1

Yes, from $p(t) + t^2\,p'(t) = 0$ it necessarily follows that $$p(t) = ke^{1/t}$$ for some $k\in\Bbb{R}$. However, the $n$-th derivative of $p$ can be calculated inductively as $$p^{(n)}(t) = ke^{1/t}\sum_{j=1}^n \frac{(-1)^j j!}{t^{j+1}} \implies p^{(n)}(1) = k \sum_{j=1}^n (-1)^jj!$$ Since $p$ is a polynomial, for large enough $n \in \Bbb{N}$ we should have $p^{(n)}(1) = 0$ but we don't even have $\lim_{n\to\infty} p^{(n)}(1) = 0$ unless $k=0$. Hence $p=0$.

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There is also a simpler solution using degree. Assume there is a $p \ne 0$ satisfying $p(t) + t^2\,p'(t) = 0$. Clearly $p$ cannot be constant so $\deg p \ge 1$. From $p(t) = -t^2p'(t)$ we get $$\deg p = 2+\deg p' = 2+(\deg p - 1) = \deg p + 1$$ which is a contradiction.