Second order non-homogeneous linear ordinary differential equation

[Hall]

I shall not begin with expressing my annoyance at the perfect equality between the number of people studying ODE and the numbers of ways of solving the Second Order Non-homogeneous Linear Ordinary Differential Equation (I'm a little doubtful about the correct syntactical position of 'linear').

I'm studying from Prof. Arthur Mattuck's lectures, and it is the lecture #13 which is confusing me. He embarks on solving (if someone who has seen my other thread can he tell me whether I have used 'embark' (an intransitive verb) in a correct way?)
$$
\begin{align*}
y^{''} + Ay^{'} + By = e^{\alpha x} &&\textrm{where α is a complex number}
\end{align*}
$$
And writes the LHS as $P(D) y$ which I understand totally. The case when $P(\alpha) \neq 0$ is clear to me, the solution is
$$
y_p = \frac{e^{\alpha x} }{P(\alpha)}
$$
But when he moves to the case when $P(\alpha) = 0$, I get a little confused. Though, I can prove
$$
\begin{align*}
P(D) e^{ax} u(x) = e^{ax} P(D+a) u(x) && \textrm{ it's no longer α, we got a there, but it is still complex}
\end{align*}
$$
Now, I don't understand things from time : 37:00 in the linked video. My doubts are thus:

• He says "if a is simple root of P(D)" then the solution is $y_p = \frac{x e^{ax} }{P'(a)}$. How can $a$ be a root of $P(D)$ which is an operator? The operator have a null space not a solution set. We should say if $a$ is a solution of polynomial $P(x)$.
• The second doubt is concerning the differentiation of $P(D)$. I still don't understand what does a derivative of an operator mean, and above that with respect to another operator $D$.

I hope my doubt shall be resolved.

 

[DrClaude]

He's talking about the polynomial $P$, which is just a polynomial. It is only when you take $P(D)$ that the entire thing becomes an operator. So simply consider what happens to the polynomial. (He should probably have written it has $P(z)$ or something like this.)

 

[Hall]

He's talking about the polynomial $P$, which is just a polynomial. It is only when you take $P(D)$ that the entire thing becomes an operator. So simply consider what happens to the polynomial. (He should probably have written it has $P(z)$ or something like this.)

Thanks for clearing that.

I'm just wondering how he guessed $y_p = \frac{x e^{ax} }{P'(a)}$? Yes, it works but how he got it?