Solving Homogeneous Linear ODEs using Annihilators - Comments

In summary, Mark44 submitted a new PF Insights post discussing the use of annihilators in solving homogeneous linear differential equations. The post explores the relationship between annihilators and characteristic equations, and how they can be used to convert nonhomogeneous differential equations into higher-order homogeneous equations. The post also addresses typos in the examples and discusses how to find the general solution without assuming a specific form for the solution.
  • #1
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Mark44 submitted a new PF Insights post

Solving Homogeneous Linear ODEs using Annihilators

ODE1-80x80.png


Continue reading the Original PF Insights Post.
 
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  • #2
I'm having a hard time seeing what the point is. It looks to me that the annihilators are the same thing as the characteristic equation.

What am I missing?
 
  • #3
ElijahRockers said:
I'm having a hard time seeing what the point is. It looks to me that the annihilators are the same thing as the characteristic equation.

What am I missing?
The differential equation in operator notation is closely related to the characteristic equation. With homogeneous diff. equations, there isn't much point, but you can use annihilators to convert some nonhomogeneous diff. equations to a higher-order homogeneous equation. That is covered in the second article, Solving Nonhomogeneous Linear ODEs using Annihilators.
 
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  • #4
Thank you very much for your artical. Please teach about "Series Solutions‐ Frobenius‘ Method".
 
  • #5
typo. 1/2i(cos(t)+isin(t))+1/2i(cos(t)–isin(t))=sin(t) has + instead of -. Should be -1/2i(cos(t)–isin(t))
 
  • #6
carllrac said:
typo. 1/2i(cos(t)+isin(t))+1/2i(cos(t)–isin(t))=sin(t) has + instead of -. Should be -1/2i(cos(t)–isin(t))
Thank you for spotting this. I have fixed this typo as well as another in that section.
 
  • #7
Hi! I am new to physicsforum. I am not sure how to post a reply/question on the article page so I am doing it here. I am also not very good at math at the moment.

My question is suppose we do not make any assumptions (that the solution is in the form y = e^(rt)), using the annihilator method, how do I find the general solution?

For example, in example 1, by assuming the solution takes the form of e^(rt), we can solve for r since r is a number. But without making this assumption, how do i get the solution y = c e^(3t) from (D-3)y = 0?
 
  • #8
zhiyang said:
Hi! I am new to physicsforum. I am not sure how to post a reply/question on the article page so I am doing it here. I am also not very good at math at the moment.

My question is suppose we do not make any assumptions (that the solution is in the form y = e^(rt)), using the annihilator method, how do I find the general solution?

For example, in example 1, by assuming the solution takes the form of e^(rt), we can solve for r since r is a number. But without making this assumption, how do i get the solution y = c e^(3t) from (D-3)y = 0?
You need to take into consideration the homogenous part of the equation. Let's take y=yc + yp, why yc is the homogenous part of the equation. You take the annihlator for the right hand side of the original equation and multiply both sides by that, giving you 0 on the right hand side and the differential operators on the left hand side. Next you find all the possible solutions for D where the right hand side would equal to 0. After you have all the values for D such that the right hand side gives you 0 you right out the solutions like you would for an ODE, and you just eliminate the solutions that are repeating from the homogenous part and keep the non repeating solutions as the "guess" to the trial function that you need for the particular solution. Thereafter you can use the left hand and use the method of undetermined co-efficients.
 
  • #9
zhiyang said:
For example, in example 1, by assuming the solution takes the form of e^(rt), we can solve for r since r is a number. But without making this assumption, how do i get the solution y = c e^(3t) from (D-3)y = 0?
If you are working with a linear, constant coefficient, homogeneous differential equation, it is reasonable to assume that the solution has the form y = cert.

(D - 3)y = 0
is the same as Dy - 3y = 0, or equivalently, y' = 3y, where y' means ##\frac{dy}{dt}##.
The DE y' = 3y is separable, and can be solved fairly easily to get the general solution: y(t) = Ce3t.
 

Related to Solving Homogeneous Linear ODEs using Annihilators - Comments

What is the concept of homogeneous linear ODEs?

Homogeneous linear ODEs are differential equations where all the terms can be expressed as a linear combination of the dependent variable and its derivatives. In other words, the dependent variable and its derivatives have a degree of 1 in each term.

What are annihilators in the context of solving homogeneous linear ODEs?

Annihilators are differential operators that are used to simplify the process of solving homogeneous linear ODEs. They are applied to the differential equation to "annihilate" certain terms, making it easier to find the general solution.

Why is using annihilators beneficial in solving homogeneous linear ODEs?

Using annihilators can greatly simplify the process of solving homogeneous linear ODEs, especially when dealing with higher order equations. They allow for a systematic approach to finding the general solution, avoiding the need for guessing and checking methods.

What are some common types of annihilators used in solving homogeneous linear ODEs?

The most commonly used annihilators are the exponential annihilator, the sine/cosine annihilator, and the binomial annihilator. These operators are used depending on the type of equation and the functions involved.

Can annihilators be used for non-homogeneous linear ODEs?

No, annihilators can only be used for homogeneous linear ODEs. For non-homogeneous equations, other methods such as variation of parameters or the method of undetermined coefficients must be used.

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