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Insights Solving Homogeneous Linear ODEs using Annihilators - Comments

  1. Dec 7, 2015 #1

    Mark44

    Staff: Mentor

  2. jcsd
  3. Dec 9, 2015 #2

    ElijahRockers

    User Avatar
    Gold Member

    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?
     
  4. Dec 9, 2015 #3

    Mark44

    Staff: Mentor

    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.
     
  5. Dec 19, 2015 #4
    Thank you very much for your artical. Please teach about "Series Solutions‐ Frobenius‘ Method".
     
  6. Dec 28, 2015 #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))
     
  7. Dec 28, 2015 #6

    Mark44

    Staff: Mentor

    Thank you for spotting this. I have fixed this typo as well as another in that section.
     
  8. Jun 9, 2016 #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?
     
  9. Jun 11, 2016 #8
    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.
     
  10. Jun 13, 2016 #9

    Mark44

    Staff: Mentor

    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.
     
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