Use the differential operator to solve this differential equation

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SUMMARY

The discussion focuses on using the differential operator method to solve the differential equation d²y/dx² - 18 dy/dx + 90y = 0. The characteristic equation method yields solutions y₁(x) = e^(9x) * cos(3x) and y₂(x) = e^(9x) * sin(3x). The differential operator approach involves expressing the equation as [D² - 18D + 90]y = 0, where D = d/dx, and factoring it into (D - r₁)(D - r₂)y = 0. By solving the first-order equations sequentially, one can derive the fundamental set of solutions.

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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!
 
Last edited:
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Since the equation is constant coefficient, the characteristic polynomial is also the differential operator form:

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

Then observer that [itex](D-r_1)\left\{ (D-r_2) y_2 \right\} = (D-r_1) {y_1} = 0[/itex] so [itex]y_2[/itex] 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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