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Differential Equations: Stuck on one step of H.L.D.E problem

  1. Mar 31, 2013 #1
    sAwgHos.jpg

    I need to find a,b,c,d, and e. I know how to do these problems the normal way but now he's giving us the answer and asking us to work backwards. I'm stumped. I think I night need to use some sord of system of equations but I'm not sure what it would look like..
     
  2. jcsd
  3. Mar 31, 2013 #2

    rude man

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    You need to perform y', y'', y''' and yiv from the given solution, substitute into the given diff. eq. then equate coefficients of like terms (in x) and solve for a, b, c and d.
     
  4. Mar 31, 2013 #3
    I did that and now I have too many variables. I have a,b,c,d,e,x,c1,c2,c3, & c4

    baih1rY.jpg
     
    Last edited: Mar 31, 2013
  5. Mar 31, 2013 #4

    rude man

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    For y', don't collect terms in e8x.
    y' = (8c2 + c4)e8x - 8c3e-8x + 8c4xe8x
    all of which is multiplied by d.
    Etc. for y, y'', etc.

    You equate coefficients of every separate term in x. So for example x*exp(x) and exp(x) are two separate terms with their own separate coefficients. Coefficients can have only the constants a, b,c,d, and e. No x's.
     
  6. Mar 31, 2013 #5
    How's that going to work though when you get down to creating a system of equations? The c1,c2,c3,c4 constants make the system unsolvable don't they?
     
  7. Apr 1, 2013 #6

    rude man

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    Why? They are given constants.
     
  8. Apr 1, 2013 #7

    HallsofIvy

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    The fact that y satisfyies the equation for any values lf [itex]c_1[/itex], [itex]c_2[/itex], etc. means that a, b, c, d, and e must be such that they cancel. Write the equation with some expression in a, b, c, d, and e times each constant and set those expressions equal to 0.
     
  9. Apr 1, 2013 #8
    So is this right?

    SwqadZ2.jpg
     
  10. Apr 1, 2013 #9
    Additionally, I can distribute the a,b,c,d,&e to all of the terms and match them all up based on their x function but that only gives me 4 equations (one for xe8x, one for e8x, e-8x, and Ec1=0). I need 5 to find the 5 variables a,b,c,d,&e. What am I missing?
     
    Last edited: Apr 1, 2013
  11. Apr 1, 2013 #10
    The system of equations I got is:

    E+8D+64C+512B+4096A=0 <--- after dividing out xe8xc4
    E+8D+64C+512B+4096A=0 <--- after dividing e8xc2
    E-8D+64C-512B+4096A=0 <--- after dividing e-8xc3
    E=0 <--- after dividing c1
    D+16c+192B+2048A=0 <--- after dividing e8xc4

    but it still doesn't solve completely because the first two equations are the same..

    you get: A=c/512, B=-c/64, C=-c/8, D=c, E=0
    If you just guess that the constant c = 1 then it's right but that's not a good system..
     
    Last edited: Apr 1, 2013
  12. Apr 1, 2013 #11

    Mark44

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    A better approach is to recognize that the characteristic equation is r(r - 8)2(r + 8) = 0. Expand this and you should be able to quickly determine the constants of your differential equation.
     
  13. Apr 1, 2013 #12

    rude man

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    They can't be the same.

    You have 5 independent equations: y,y',y'',y''' and y''''. And 5 unknowns a,b,c,d and e.

    It also looks like you have not segregated the exp(x) and the xexp(x) terms as I pointed out you should. And how did you wind up with an exp(16x) in your 6th line from the bottom on post 8? Not likely in a linear diff. eq.!

    Let's take a simpler example:
    y = c1 + c2exp(x)+ c3xexp(x) is the solution to
    ay' + by + c = 0.
    Then y' = c2exp(x) + c3exp(x) + c3xexp(x)

    So substitute ino the diff eq:
    a(c2 + c3) exp(x) + ac3 xexp(x)] + bc1 + bc2 exp(x) + bc3 xexp(x)] + c = 0

    from which we group coefficients of like terms in x including x0:

    a(c2 + c3) + bc2 = 0 groups the exp(x) coeff.
    ac3 + bc3 = 0 groups the x exp(x) coeff.
    bc1 + c = 0 groups the exp(0) coeff.

    & solve for a, b and c.
    See now why you should not have included an x in your coefficients?
     
  14. Apr 1, 2013 #13
    When you do that you get:
    r4-8r3-64r2+512x
    What would the systematic next step be because the real answer seems to be A=1/512, B=-1/64, C=-1/8, D=1, E=0
     
  15. Apr 1, 2013 #14

    Mark44

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    You get r4 - 8r3 - 64r2 + 512r = 0, so the DE would be y(4) - 8y''' - 64y'' + 512y' + 0y = 0.

    [STRIKE]Now that the coefficients of the DE are known, you can find the coefficients of the solution.[/STRIKE]
     
    Last edited: Apr 1, 2013
  16. Apr 1, 2013 #15
    I've gone over it several times now. I'm following the same steps as you can see in the work below but I don't see *specifically* where I went wrong. I ultimately get:
    factoring out c4xe8x --> 4096A+512B+64C+8D+E=0
    factoring out c4e8x --> 2048A+192B+16C+D=0
    factoring out c2e8x --> 4096A+512B+64C+8D+E=0
    factoring out c3e-8x --> 4096A-512B+64C-8D+E=0
    factoring out c1 --> E=0
     
  17. Apr 1, 2013 #16
    Wait, I don't follow. The question asks for a,b,c,d,&e of the DE. If you're saying that A=1,B=-8,C=-64, D=512, and E=0 then I'm done right? If that's true, the other solutions I found for a,b,c,d,&e also work so are they both right?

    Additionally, I was under the impression that the constants for the solution can not be found without boundary values yet you're implying that they can be.
     
  18. Apr 1, 2013 #17

    Mark44

    Staff: Mentor

    Yes - that's all they're asking for.
    I think so. I haven't checked closely, but your constants seem to be a multiple of the ones I found.

    Edit: Yes, my numbers are 512 times yours.
    The constants of the solution (c1, etc.) can't be found without initial conditions. I erred in saying that in my previous post, so I will take it out.
     
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