# Second order non-homogeneous linear ordinary differential equation

• I
• Hall
In summary, the conversation is about solving a Second Order Non-homogeneous Linear Ordinary Differential Equation with a complex number alpha. The speaker is confused about the correct syntactical position of 'linear' and the usage of the word 'embark'. They are studying from Prof. Arthur Mattuck's lectures and are confused about lecture #13. The lecturer uses the operator P(D) to represent the left hand side of the equation and explains the solution when P(alpha) is not equal to 0 and when it is equal to 0. The speaker has doubts about a being a simple root of P(D) and the differentiation of P(D). The expert summarizer resolves the doubts by clarifying that a is a root of the polynomial P,

#### 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.

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 and topsquark
DrClaude said:
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?