Use the differential operator to solve this differential equation

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The discussion focuses on using the differential operator method to solve the second-order differential equation d^2y/dx^2 - 18dy/dx + 90y = 0. The characteristic equation is derived as r^2 - 18r + 90 = 0, yielding complex roots r = 9 ± 3i, leading to solutions y_1(x) = e^(9x) * cos(3x) and y_2(x) = e^(9x) * sin(3x). The differential operator form is expressed as [D^2 - 18D + 90]y = 0, where D represents differentiation. By factoring the characteristic polynomial, the method involves solving first-order equations to derive the fundamental set of solutions. Understanding this approach is crucial for mastering the differential operator method in solving such equations.
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Homework Statement


Use the differential operator method to find a fundamental set of solutions {y_1(x), y_2(x)} of the equation

d^2 y/dx^2 - 18 dy/dx + 90y = 0.


Homework Equations


Differential operator method.


The Attempt at a Solution


I have a huge problem understanding how to use the differential operator method. I can successfully complete this problem using the characteristic equation:

r^2 - 18r + 90 = 0
r = 9 +/- 3i
y_1(x) = e^(9x) * cos(3x)
y_2(x) = e^(9x) * sin(3x)

but I really need to understand how to use the differential operator and I don't get anything I read on the internet or in my textbook.

A comparison (with contrasting) to the method with the characteristic equation would be GREATLY appreciated!
Thanks in advance!
 
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Since the equation is constant coefficient, the characteristic polynomial is also the differential operator form:

Your equation takes the form:
[D^2 -18D +90]y = 0, where D = \frac{d}{dx}
Factoring the characteristic polynomial becomes a factorization of the differential operator:
(D-r_1)(D-r_2)y = 0
you can then successively integrate:
Solve the first order equation: (D-r_1)y=0 and let y_1 be your general solution.
Then solve the first order equation (D-r_2)y = y_1 and let that solution be y_2

Then observer that (D-r_1)\left\{ (D-r_2) y_2 \right\} = (D-r_1) {y_1} = 0 so y_2 is a solution to your original equation. If you gave the most general solution in each step you have the most general solution to the original problem.
 
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