The Method of Differential Annihilators

The Method of Differential Annihilators

Now that we have looked at Differential Annihilators, we are ready to look into The Method of Differential Annihilators. Once again, this method will give us another way to solve many higher order linear differential equations as opposed to the method of undetermined coefficients.

Suppose that $L(D)$ is a linear differential operator with constant coefficients and that $g(t)$ is a function containing polynomials, sines/cosines, or exponential functions. Then this method works perfectly for solving the differential equation:

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

We begin by solving the corresponding linear homogenous differential equation $L(D)(y) = 0$. We then determine a differential operator $M(D)$ such that $M(D)(g(t)) = 0$, that is, $M(D)$ annihilates $g(t)$. Then we apply this differential operator to both sides of the differential equation above to get:

(2)
\begin{align} \quad M(D)L(D)(y) = M(D)(g(t)) \\ \quad M(D)L(D)(y) = 0 \end{align}

We thus obtain a linear homogenous differential equation with constant coefficients, $M(D)L(D)(y) = 0$. We can then easily solve this differential equation. We will get a general solution to $M(D)L(D)(y) = 0$. To get a particular solution to $L(D)(y) = g(t)$, we will eliminate terms that are linear combinations of the general solution corresponding linear homogenous differential equation $L(D)(y) = 0$.

The terms that remain will be of the appropriate form for particular solutions to $L(D)(y) = g(t)$. We then plug this form into this differential equation and solve for the values of the coefficients to obtain a particular solution.

Perhaps the method of differential annihilators is best described with an example. Consider the following third order differential equation:

(3)
\begin{align} \quad \frac{d^3y}{dt^3} + 6 \frac{d^2y}{dt^2} + 11 \frac{dy}{dt} + 6y = 2e^t + e^{-t} \end{align}

Note that this is a third order linear nonhomogenous differential equation, and the function $g(t) = 2e^t + e^{-t}$ on the right hand side of this differential equation is in a suitable form to use the method of undetermined coefficients. Once again we'll note that the characteristic equation for this differential equation is:

(4)
\begin{align} \quad r^3 + 6r^2 + 11r + 6 = 0 \end{align}

This characteristic equation can be nicely factored as:

(5)
\begin{align} \quad (r + 1)(r + 2)(r + 3) = 0 \end{align}

Thus we get the general solution to our corresponding third order linear homogenous differential equation is $y_h(t) = Ae^{-t} + Be^{-2t} + Ce^{-3t}$.

Now we can rewrite our original differential equation in terms of differential operators that match this characteristic equation exactly:

(6)
\begin{align} \quad (D + 1)(D + 2)(D + 3)y = 2e^t + e^{-t} \end{align}

Now note that $(D - 1)$ is a differential annihilator of the term $2e^t$ since $(D - 1)(2e^t) = D(2e^{t}) - (2e^{t}) = 2e^t - 2e^t = 0$. Furthermore, note that $(D + 1)$ is a differential annihilator of the term $e^{-t}$ since $(D + 1)(e^{-t}) = D(e^{-t}) + (e^{-t}) = -e^{-t} + e^{-t} = 0$. Note that also, $(D - 1)(D + 1)(-e^{-t} + e^{-t}) = (D^2 - 1)(-e^{-t} + e^{-t}) = D^2(-e^{-t} + e^{-t}) - (-e^{-t} + e^{-t}) = -e^{-t} + e^{-t} + e^{-t} - e^{-t} = 0$.

We will now apply both of these differential operators, $(D - 1)(D + 1)$ to both sides of the equation above to get:

(7)
\begin{align} \quad (D - 1)(D + 1)^2(D + 2)(D + 3)y = (D - 1)(D + 1)(2e^t + e^{-t}) \\ \quad (D - 1)(D + 1)^2(D + 2)(D + 3)y = 0 \\ \quad (D^2 - 1)(D^3 + 6D^2 + 11D + 6)y = 0 \\ \quad (D^5 + 6D^4 + 11D^3 + 6D^2 - D^3 - 6D^2 - 11D - 6)y = 0 \\ \quad (D^5 + 6D^4 + 10D^3 - 11D - 6)y = 0 \\ \quad \frac{d^5y}{dt^5} + 6 \frac{d^4y}{dt^4} + 10 \frac{d^3y}{dt^3} - 11 \frac{dy}{dt} - 6y = 0 \end{align}

Thus we have that $y$ is a solution to the homogenous differential equation above. Note that the corresponding characteristic equation is given by:

(8)
$$r^5 + 6r^4 + 10r^3 - 11r - 6 = 0$$

The roots to the characteristic polynomial are actually given by the factored form of the polynomial of differential operators from earlier, and $r_1 = 1$, $r_2 = -1$ (with multiplicity 2), $r_3 = -2$, and $r_4 = -3$, and so for some constants $D$, $E$, $F$, $G$, and $H$ we have that:

(9)
\begin{align} \quad y = De^{t} + Ee^{-t} + Fte^{-t} + Ge^{-2t} + He^{-3t} \end{align}

Note that the terms $Ee^{-t}$, $Ge^{-2t}$, and $He^{-3t}$ form a linear combination of the solution to our corresponding third order linear homogenous differential equation from earlier, and so we can dispense with them in trying to find a particular solution for the nonhomogenous differential equation, so $y = De^t + Fte^{-t}$. We will now differentiate this function three times and substitute it back into our original differential equation. We have that:

(10)
\begin{align} \quad \frac{dy}{dt} = De^t + Fe^{-t} - Fte^{-t} \end{align}
(11)
\begin{align} \quad \frac{d^2y}{dt^2} = De^{t} -Fe^{-t} - (Fe^{-t} - Fte^{-t}) \\ \quad \frac{d^2y}{dt^2} = De^{t} -2Fe^{-t} + Fte^{-t} \end{align}
(12)
\begin{align} \quad \frac{d^3y}{dt^3} = De^{t} + 2Fe^{-t} + (Fe^{-t} - Fte^{-t}) \\ \quad \frac{d^3y}{dt^3} = De^{t} + 3Fe^{-t} - Fte^{-t} \end{align}

Plugging these into our third order linear nonhomogenous differential equation and we get that:

(13)
\begin{align} \quad (De^{t} + 3Fe^{-t} - Fte^{-t}) + 6(De^{t} -2Fe^{-t} + Fte^{-t}) + 11(De^t + Fe^{-t} - Fte^{-t}) + 6(De^t + Fte^{-t}) = 2e^t + e^{-t} \\ \quad 24De^t + 2Fe^{-t} = 2e^t + e^{-t} \end{align}

The equation above implies that $D = \frac{1}{12}$ and $F = \frac{1}{2}$, and so a particular solution to our third order linear nonhomogenous differential equation is $y_p = \frac{1}{12}e^t + \frac{1}{2} t e^{-t}$, and so the general solution to our differential equation is:

(14)
\begin{align} \quad y = Ae^{-t} + Be^{-2t} + Ce^{-3t} + \frac{1}{12}e^t + \frac{1}{2} t e^{-t} \end{align}