Simmultanous equations

Discussion in 'Math' started by jagjit Sehra, Mar 4, 2008.

  1. jagjit Sehra

    Thread Starter Member

    Feb 29, 2008
    25
    0
    Hi Folks,
    Got a question regarding simmultanous equations. I am studying Electronics and have been looing at a certain type of response. The crtically damped response.

    vC = Aexp(mt) + Btexp(mt)
    where m is one value lets say -1. vC has intial

    value of 2.

    The problem is to obtain the values for A and B.

    The derivative of vC with respect time.

    dvC/dt = mAexp(mt) + Bexp(mt) + mBtexp(mt)

    (differentiated the second term using the product rule)

    assuming t = 0
    Putting values in.

    2 = Aemt + Btemt
    0 = mAemt + Bemt + mBtemt

    2 = A + 0 A must equal 2
    0 = -A + B + 0

    0 = -2 + B B must equal 2

    Therefore A and B both equal 2. Is that correct the way I have done it! It doesnt look right!
     
  2. Mark44

    Well-Known Member

    Nov 26, 2007
    626
    1
    It's not clear to me what you're doing. At the beginning you say assume that m = -1, but you don't use that value in your later work.

    After taking the derivative, it looks like you are setting d(VC)/dt to 0, but you don't mention that this is given information.

    Using all the information that you have, does A = B = 2 make sense? That is, do these values satisfy both the original equation and its derivative when t=0, and presumably for m=-1?

    Mark
     
  3. Mark44

    Well-Known Member

    Nov 26, 2007
    626
    1
    If you have these equations:

    V(t) = A*exp(-t) + B*t*exp(-t)
    V'(t) = -A*exp(-t) -B*t*exp(-t) + B*exp(-t)

    with these initial conditions:
    V(0) = 2, V'(0) = 0

    Then, yes, your work is correct; namely, that A = B = 2.

    I used your assumed value of m = 1.
    Mark
     
  4. jagjit Sehra

    Thread Starter Member

    Feb 29, 2008
    25
    0
    Thanks Mark for confirming that. I must also apologise for bieng so vague with the description of the problem. I must say I have found this foroumn very usefull and hopefully one day day I can help somone aswell!
     
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