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Diff eq please help me!

  1. Oct 13, 2008 #1
    Find the value of the constant c so that solutions of the equation 4u"+cu'+6u=0 tend to zero as fast as possible.


    I think that c must be positive in order for this to tend towards zero but I cannot figure out what c has to be.


    Thanks for your help!
     
  2. jcsd
  3. Oct 13, 2008 #2

    gabbagabbahey

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    What is the characteristic polynomial of this ODE? What are the solutions (in terms of c) of that polynomial? What does that make the general solution for u?
     
  4. Oct 13, 2008 #3
    this is all i was given.
    would it be that m=4 lambda=c and k=6?
     
  5. Oct 13, 2008 #4

    gabbagabbahey

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    Have you studied ODE's yet?? Are you familiar with the term characteristic polynomial?

    For example, the characteristic polynomial of [itex]2u''+5u'-3=0[/itex] is [itex] 2 \lambda^2+5\lambda-3=0[/itex] which has roots [itex] \lambda_1= -3[/itex] and [itex]\lambda_2=\frac{1}{2}[/itex] and a general solution of [itex]u(x)=c_1e^{\lambda_1 x} +c_2e^{\lambda_2 x}[/itex]

    So, what is the characteristic polynomial of 4u" + cu'+6u=0?
     
  6. Oct 13, 2008 #5
    So the equation would be [itex] 4\lambda^2+C\lambda+6=0[/itex]
     
  7. Oct 13, 2008 #6
    and the roots would be (-C+/- the square root(c^2 - 96))/8
     
  8. Oct 13, 2008 #7

    gabbagabbahey

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    Yes, so what is the general solution u(x) then?
     
  9. Oct 13, 2008 #8
    u(x) = c1e^(-C + the square root(c^2 - 96))/8)x + C2e^(-C - the square root(c^2 - 96))/8
     
  10. Oct 13, 2008 #9

    gabbagabbahey

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    Good, now what does it mean for a function to tend to zero? (in terms of the derivative of the function)
     
  11. Oct 13, 2008 #10
    its going to have to be getting infinitely smaller
     
  12. Oct 13, 2008 #11

    gabbagabbahey

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    Yes, it gets smaller as x-gets larger and so du/dx is negative correct?

    What is du/dx of your function?
     
  13. Oct 13, 2008 #12
    im not sure how to find that here. is that the derivative in regards to x?
     
  14. Oct 13, 2008 #13
    im really stuck. would it be that C must b be 1 to tend towards zero?
     
  15. Oct 13, 2008 #14

    gabbagabbahey

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    Yes, du/dx is the derivative with regard to x...You know what u(x) is, so calculate u'.
     
    Last edited: Oct 13, 2008
  16. Oct 13, 2008 #15
    okay i did that. now where in there can i find what c is?
     
  17. Oct 13, 2008 #16

    gabbagabbahey

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    Well u' should be as large of a negative number as possible if u tends to zero as fast as possible correct?

    In other words you want to find the minumum value of u'(x) with respect to c. The derivative of u'(x) with respect to c should be zero at its minimum, so solve the equation

    [tex]\frac{d u'}{dc}=0[/tex]

    for c.
     
  18. Oct 13, 2008 #17
    "(the solution) tends to zero as fast as possible"

    My interpretation is that the transient response is "as short as possible".
    given a second order system [tex]ay''+by'+cy=dx'+ex[/tex] under zero input excitation or [tex]ay''+by'+cy=0[/tex], the response is due to non-zero initial condition(s). In order for the solution to converge (stable system and non oscillating), it requires either
    1) a>0,b>0, and c>0, OR
    2) a<0, b<0, and c<0
    such that all poles are on the left half of s-plane (excluding [tex]j \omega[/tex]) axis. In the other word, the fourier transform must exist.

    since the OP mentioned m, c, and k, I'm assuming she is dealing with a spring-mass-damper system ([tex]my''+cy'+ky=0[/tex] with non-equilibrium initial displacement
    ), I bet the answer is when [tex]\zeta=1[/tex] or critically damped. Note that when system is critically damped (double real roots), the solution doesn't take the form as other poster mentioned.
     
    Last edited: Oct 13, 2008
  19. Oct 13, 2008 #18

    gabbagabbahey

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    Yes, that makes more sense :0)
     
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