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Homework Help: Homogenous Solution Represents the Transient Response Right?

  1. Mar 14, 2016 #1

    Does the Homogenous Solution represent the Transient Response?

    Let me specify. For a N-DOF spring, mass, and damper mechanical system:

    -Does the Homogenous Solution represent the Transient Response for given mechanical system?



    -only stable systems are considered

    According to Swarthmore University [Link listed]:


    Finding the homogeneous and particular solutions is a general technique for solving differential equations of the sort that we will encounter (different inputs require different forms of the particular response, but we will only consider step inputs (i.e., the input is constant for t>0) for now. However, contemplation of the technique begs the question: What, physically, do the homogeneous and particular response represent. The particular response represents the response of the system after any initial transients have died out, but the the homogeneous response doesn't really represent anything physical. The reason we use it is that it is mathematically correct and yields the right answer.

    END QUOTE. (You can search that and Cltrl F (Command F) and paste that quote to find where this is listed)

    Please note:

    If their statement is only applicable to electrical systems. Then can you tell me how?
  2. jcsd
  3. Mar 14, 2016 #2


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    I don't think you should let them confuse you. Swarthmore isn't CalTech or MIT ...
    But they do have a nice website :smile:
  4. Mar 14, 2016 #3
    Ok. Yes it is interesting.

    Thank you for the reply.
  5. Mar 17, 2016 #4
    So, I was beginning to build a new and deeper understanding when I noticed that Particular Solutions can also be transient:
    Thus in that scenario, I believe, the Particular Solutions can, jointly, represent the Transient Response. And as expected there isn't a Steady State Response because the system returns to equilibrium, so the Total Response is transient because it is the Transient Response.

    Then.....I was presented with an example (via email) from a Professor at Swarthmore (listed below);
    It basically counters my new understanding that the Homogenous Solution, completely or at least jointly, represent the Transient Response.

    -----------------------------------------------------------------------------------Begin Math---------------------------------

    Example: dy/dt + y = exp(-t) ; Initial Conditions: y(t=0) = 0, & dy/dt(t=0) = 0
    To find the total solution I tried two techniques: 1) Integrating Factors 2) Summing Complementary (Homogenous) and Particular Solutions.
    1) Integrating Factors
    mu = exp(int(coefficient of y*dt)) = exp(int(1*dt)) = exp(t)
    multiply ODE by mu:
    mu*dy/dt + mu*y = mu*exp(-t) = exp(t)*exp(-t) = 1
    noticing product rule:
    d/dt*[exp(t)*y] = 1
    integrating both sides:
    y*exp(t) = t+C
    Initial Conditions:
    y(t) = exp(-t)*(t+C) => y(0) = 0 = exp(-0)*(0+C) => C = 0
    Total Solution by Integrating Factor Technique:
    y(t) = t*exp(-t)
    2) Sum of Complementary (Homogenous) Solution and Particular Solution
    Complementary (Homogenous) Solution: set right hand side of ODE equal to zero, and then can just do typical integration or can assume solution to be yh = A*exp(t)
    I chose to do the typical integration:

    dy/dt + y = 0 => dy/dt = -y => 1/y *dy = -dt
    log|y| +C1 = -t log here is natural logarithm
    exp(log|y|) = exp(-t-C1) => yh(t) = exp(-t)*exp(-C1); if you did this by assumption/guess then your A would be equal to the exponential of C1
    Particular Solution: referred to textbook for this one because if I assume a particular solution of the form of that similar to RHS of ODE yp(t) = B*exp(-t) then I duplicate the complementary/homogenous solution i.e.: dy/dt + y = -Aexp(-t) + A*exp(-t) = exp(-t) = A*[-exp(-t)+exp(-t)] = exp(-t) = A[0] = exp(-t) => cannot determine A

    Instead assuming: yp(t) = B*t*exp(-t); by reasoning above
    dy/dt + y = exp(-t) = Aexp(-t)-A*t*exp(-t)+A*t*exp(-t) = exp(-t) = A*exp(-t) = exp(-t) => A = 1 => yp(t) = t*exp(-t) due to assumed solution [different than just B*exp(-t)]
    Apply Initial Conditions to Total Solution:
    y(t) = yh(t) + yp(t) = exp(-t)*exp(-C1) + t*exp(-t) => y(0) = 0 = exp(-0)*exp(-C1) +0*exp(-0) => exp(-C1) = 0 => C1 = infinitely large number that drives exponent to zero
    y(t) = exp(-t)*exp(-infinity) + t*exp(-t) = exp(-t)*0 + t*exp(-t) = 0 + t*exp(-t); => yh(t) = 0 due to zero initial conditions
    Total Solution by Complementary (Homogenous) Solution and Particular Solution :
    y(t) = 0 + t*exp(-t);
    notice both methods result in same

    --------------------------------------------------------------------------------------End Math---------------------------------

    Questions: (3)
    1) I think that the Homogenous Solution does not always solely represent the Transient Response. What's more, if the Homogenous Solution is zero then it does not represent any part of the Transient Response (and it should go without saying that it doesn't represent any part of the total response)
    So, to keep it general: the Homogenous Solution, if non-zero, typically and solely represents the Transient Response provided the Particular Solution isn't transient.
    2) How can there, physically, be a particular solution without a homogenous solution? Is it just because, mathematically, it's due to zero initial conditions? Does this occasionally or quite often happen?
    3) Also, what real life examples has an input (forcing function?) that is exponentially decreasing with time or rather a particular solution that represents, or partially represents (if the homogenous solution is non zero) the Transient Response?

    I think the example is misleading because it sort of makes one think that if the Particular Solution is transient then the Homogenous Solution must be zero.
    That's definitely not always the case.

    You can have, at least mathematically, a Particular Solution that is transient and likewise a Homogenous Solution that is non-zero and thus both solutions together represent the Transient Response. (Again there isn't a Steady State Response here, and thus the Total Response is the Transient Response; only response )
    Last edited: Mar 17, 2016
  6. Mar 17, 2016 #5


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    I think you have it all figured out prety well -- and I stand corrected for sure. Must be the usual physicists sloppiness taking the usual case for always true. In defence: transient particular solutions (such as step response) are easily singled out and often dealt with using other methods.
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