# 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)

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)

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)

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)

This characteristic equation can be nicely factored as:

(5)

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)

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)

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

(8)

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)

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)

(11)

(12)

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

(13)

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)

### Related post:

- Grade Nine learners taught mathematics skills – Tembisan
- A Library Browse Leads Math’s Bill Dunham to Question the Origins of The Möbius Function – Bryn Mawr Now
- Year 5 and 6 students to sit competition this Wednesday – Great Lakes Advocate
- USC student wins silver medal in China math contest – SunStar Philippines
- CBSE Exam 2020: Two separate examinations to be conducted for Class 10 Mathematics – Jagran Josh
- Concepts incomplete, problems unsolvable in math textbooks – Times of India
- Education Ministry to Host Tertiary and Employment Fairs – Government of Jamaica, Jamaica Information Service
- Vogue’s Edwina McCann and Westpac’s Anastasia Cammaroto on how they inspire women to pursue STEM – Vogue Australia
- Jonee Wilson, Temple Walkowiak to Measure High-Quality Instructional Practices to Support Marginalized Students in Rigorous Mathematics through NSF Grant – NC State College of Education
- Australian Conference on Science and Mathematics Education – Australian Academy of Science