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Solve 12y'' -5y' -2y = 0

  1. Oct 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve 12y'' -5y' -2y = 0

    2. Relevant equations



    3. The attempt at a solution

    So I'm not really sure how to do this at all. I feel like he took this out of left field.

    What is I rewrote it as 12y'' -5y' = 2y Then what if I integrated it.

    I may get 12y' - 5y = y + C ? Then I could use an integrating factor because it is of that form? And just go from there? Is this OK. Or am I making stuff up?
     
  2. jcsd
  3. Oct 30, 2013 #2

    Dick

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    Making stuff up. If you integrate 2ydy you get y, but you need to integrate 2y with respect to whatever the independent variable is. And it's not y. This is a different type of problem. It's a linear differential equation with constant coefficients. http://en.wikipedia.org/wiki/Linear_differential_equation
     
  4. Oct 30, 2013 #3
    It sent me to nothing .... I got this Wikipedia does not have an article with this exact name

    How to I handle this problem
     
  5. Oct 30, 2013 #4
    I regoogled it and found the article idk why ur link didn't work
     
  6. Oct 30, 2013 #5

    Dick

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    Because my copy and paste function often double copies and I missed that one. So it's badly formatted. I fixed it.
     
  7. Oct 30, 2013 #6
    Sorry I'm still unsure of how to go about this problem. I'm looking over the example. But I don't understand really.
     
  8. Oct 30, 2013 #7

    Dick

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    You probably just haven't looked at it enough. Assuming your dependent variable is x, assume a solution has the form y=e^(kx). Put that into the ode and try to figure out what the possibilities for k are. That's the jist of it. It's pretty simple really.
     
  9. Oct 30, 2013 #8
    OK. I did it a little different looking at an example. I just

    let y'' = r^2 and y' = r in my equation

    12r^2 -5r - 2 = 0 and I factored it. I get r = (1/4) and r = (2/3) so then I looked at how this example reported there answer and they had.

    Y(x) = C1 e^(1/4)z + C2e^(2/3)z

    I'm not sure why they choose z or even why it get;s this form for the solution
     
  10. Oct 30, 2013 #9
    I guess I should use x
     
  11. Oct 30, 2013 #10

    Dick

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    You are trying to go too fast without understanding what you are doing. For one thing, I get r=(-1/4) (not +1/4) or r=(2/3). If your independent variable is x then e^(-x/4) and e^(2x/3) are both solutions. Check that first.
     
  12. Oct 30, 2013 #11
    Well I'm going fast because it is late here an I have examination in another course in the morning so forgive me for my lack of understanding. So, you don't even get 2/3 as an answer?
     
  13. Oct 30, 2013 #12

    Dick

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    Yes, I do. Didn't I say that I did? Why are asking? This is what I mean by 'too fast'.
     
  14. Oct 30, 2013 #13
    Kind of I got confused by your language. So I should pop this those in my original differential equation? Yes I think so.
     
  15. Oct 30, 2013 #14
    Alright I got 0 for both in the original. So what is with how they write the answer? In that form?
     
  16. Oct 30, 2013 #15

    Dick

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    Good idea. That should make it doubly clearer why the roots of 12r^2 -5r - 2 = 0 give you the exponents for solutions to your ODE.
     
  17. Oct 30, 2013 #16
    OK so what about this one because I can't factor this like the other one....

    y'' -4y' +5y = 0
     
  18. Oct 30, 2013 #17

    Dick

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    They should have written y(x)=C1*e^(-x/4)+C2*exp(2x/3), with C1 and C2 being constants. I don't know where the z came from. You can see that also solves your diffy q, right? The rest of the story, which I'm not going to go into here, is that a 2nd order ode like that has exactly two independent linear solutions. So that is not only A solution, ALL of the solutions must have that form. I'm sure all of this stuff is in your textbook or course somewhere.
     
  19. Oct 30, 2013 #18

    Dick

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    You can still find complex roots which will give you complex exponentials. Which will turn out to be products of sines, cosines and ordinary real exponentials via Euler's formula. This stuff has got to be in your course someplace! I'm not going to try to explain the whole thing from scratch.
     
  20. Oct 30, 2013 #19
    y(x)=C1*e^(-x/4)+C2*exp(2x/3)

    What is exp(2x/3)?
     
  21. Oct 30, 2013 #20

    Dick

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    e^(2x/3). I sometimes write e^x as exp(x). I slipped.
     
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