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**Chapter 4 / A clarification on the method of annihilators**

« **on:**October 29, 2019, 02:13:16 PM »

Hi TUT0101 (and everyone else),

I just wanted to clear some things up regarding our discussion of the method of annihilators today (October 29) in tutorial. The question was: why does the method of annihilators always lead to the correct guess for a particular solution.

Consider the non-homogeneous linear ODE

\begin{align} \label{1} L(D)y = g(t). \end{align}

First, find an operator $H(D)$ which annihilates $g$. That is, find $H(D)$ so that

\begin{align} \label{2} H(D)g = 0.\end{align}

The choice of $H(D)$ is highly sensitive to $g$ itself, and it is not always possible to quickly find $H(D)$. If $g$ is an exponential, trigonometric function (sine or cosine), polynomial, or some products of these, then it is possible to find $H(D)$. Applying $H(D)$ to \eqref{1} gives

\begin{align}\label{3} H(D)L(D)y = H(D)g = 0.\end{align}

We can write the general solution to the ODE \eqref{3} as $c_1y_1 + \cdots + c_ny_n$, where $y_1, \ldots, y_n$ is some fundamental set of solutions. But we can also get a solution of \eqref{3} by looking at \eqref{1}: if $y_p$ is some particular solution to \eqref{1}, i.e. $L(D)y_p = g(t)$, then $y_p$ is also a solution to \eqref{3}, since \begin{align} H(D)L(D)y_p = H(D)g = 0.\end{align}

So then $y_p$ must be some linear combination of the fundamental solutions $y_1, \ldots, y_n$. That is, there is some correct choice of constants so that $y_p = c_1y_1 + \cdots c_ny_n$. Of course, if it turns out $y_i$ solves $L(D)y_i = 0$, then it does not help solve $L(D)y_p = g(t)$, so we may take the corresponding $c_i$ to be $0$.

Therefore particular solutions to \eqref{1} are always linear combinations of fundamental solutions to \eqref{3}, which means the method of annihilators always leads to the correct guess of the form of the particular solution to \eqref{1}.

David

I just wanted to clear some things up regarding our discussion of the method of annihilators today (October 29) in tutorial. The question was: why does the method of annihilators always lead to the correct guess for a particular solution.

Consider the non-homogeneous linear ODE

\begin{align} \label{1} L(D)y = g(t). \end{align}

First, find an operator $H(D)$ which annihilates $g$. That is, find $H(D)$ so that

\begin{align} \label{2} H(D)g = 0.\end{align}

The choice of $H(D)$ is highly sensitive to $g$ itself, and it is not always possible to quickly find $H(D)$. If $g$ is an exponential, trigonometric function (sine or cosine), polynomial, or some products of these, then it is possible to find $H(D)$. Applying $H(D)$ to \eqref{1} gives

\begin{align}\label{3} H(D)L(D)y = H(D)g = 0.\end{align}

We can write the general solution to the ODE \eqref{3} as $c_1y_1 + \cdots + c_ny_n$, where $y_1, \ldots, y_n$ is some fundamental set of solutions. But we can also get a solution of \eqref{3} by looking at \eqref{1}: if $y_p$ is some particular solution to \eqref{1}, i.e. $L(D)y_p = g(t)$, then $y_p$ is also a solution to \eqref{3}, since \begin{align} H(D)L(D)y_p = H(D)g = 0.\end{align}

So then $y_p$ must be some linear combination of the fundamental solutions $y_1, \ldots, y_n$. That is, there is some correct choice of constants so that $y_p = c_1y_1 + \cdots c_ny_n$. Of course, if it turns out $y_i$ solves $L(D)y_i = 0$, then it does not help solve $L(D)y_p = g(t)$, so we may take the corresponding $c_i$ to be $0$.

Therefore particular solutions to \eqref{1} are always linear combinations of fundamental solutions to \eqref{3}, which means the method of annihilators always leads to the correct guess of the form of the particular solution to \eqref{1}.

David