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Homework Help: ODEs: Why do I need Both a General And a Particular Solution?

  1. Jun 6, 2006 #1
    I am asking this question as it relates to physics, and in particular how it relates to harmonic oscillation.

    Why is the equation not solved when I use only a particular solution? Why is the equation not solved when I use only a general solution?
     
  2. jcsd
  3. Jun 6, 2006 #2

    Hootenanny

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    The obvious thing to say would be that a particular solution is only valid in a specific case, where as a general solution is always valid but is dependent upon intital conditions.
     
    Last edited: Jun 6, 2006
  4. Jun 6, 2006 #3

    AKG

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    You want to be able to describe all possible solutions. When someone asks you to solve x2 = 1, you give all possible solutions: 1 and -1. So when someone asks you to solve and ODE, you want to give all possible solutions. If you take the general solution (to the related homogenous equation) and add it to a particular solution, you are able to describe all possible solutions. Consider:

    y'' - 5y' + 6y = 3

    The related homogenous equation is:

    y'' - 5y' + 6y = 0

    which has general solution

    yH(x) = ce2x + de3x

    Note the above is really a family of solutions depending on the paramters c and d. A particular solution to the original equation is:

    yP(x) = 1/2

    So the solution to the original equation is

    y(x) = ce2x + de3x + 1/2

    Again, this is a family of solutions depending on paramters c and d. This fully describes the set of solutions to the original ODE. By "fully" describes I mean that:

    a) if you replace c and d with any two real numbers, you will get a real function that solves the original ODE, i.e. IF y(x) = ce2x + de3x + 1/2 for some c and d THEN y(x) solves the original ODE
    b) you can get EVERY solution to the ODE by plugging in numbers for c and d, i.e. IF y(x) solves the original ODE THEN y(x) = ce2x + de3x + 1/2 for some c and d.
     
  5. Jun 6, 2006 #4
    That is not correct. A general solution is as much of a solution as a particular solution. Its just that a general solution does not incorporate initial conditions whereas a particular solution does. So for a physical application it is necessary to place restrictions imposed by physics on a solution obtained by solving the d.e. to restrict its class and to make sure it gives physically meaningful results.
     
  6. Jun 8, 2006 #5
    The equation is solved if you present a particular solution, simply because you are giving a certain function that agrees with the D.E. Your homework problem might not be solved by only giving 1 particular solution, however to say that the DE is not solved is incorrect.
     
  7. Jun 8, 2006 #6
    The particular solution often times solves the equation for the forcing function, while the homogeneous solution is the steady-state solution of the equation. if you are dealing with homogenous ode, the particular solution and the homogeneous solution are one in the same.

    so everyone is right partially:

    the particualar solution is the non-equilibrium solution

    please make sure to add the particular and homogeneous solutions together before applying boundary conditions.
     
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