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Why particular solutions?

  1. Sep 28, 2015 #1
    If the purpose of the general form of the solution to a differential equation is to represent a formula with parameters for the solutions to that differential equation, why is it that we typically want to add some particular solution to the general one?

    Solution = General Solution + Particular Solution.

    I suppose I understand why you'd want it there, but the part that I don't remember from my Diff. Eq. stuff from long ago is why the particular solution is not included in the general solution.
  2. jcsd
  3. Sep 28, 2015 #2


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    Gold Member

    The particular solution is for the corresponding non-homogeneous equation. In other words, on the right hand side of the differential equation, zero is replaced with some function.
  4. Sep 28, 2015 #3


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    A particular solution is a solution that satisfies boundary or initial value conditions. It shows up in inhomogeneous differential equations, as far as I recall.
  5. Sep 28, 2015 #4
    So, is this appropriate?

    The general solution for a non-homogeneous system of differential equations is analogous to the +C of integration. Because of the superposition principle, any function of the form of the general solution COULD be a part of the solution for the ODE since it would have become exactly zero upon substitution for y.
  6. Sep 28, 2015 #5


    Staff: Mentor

    No, this isn't quite right. It is General Solution = Complementary solution + Particular Solution = yc + yp. The complementary solution is the solution to the associated homogeneous problem.
    The homogeneous problem is f(t, y, y', ... ,y(n)) = 0; for example, y'' + 4y' + 4y = 0.
    The nonhomogeneous problem is f(t, y, y', ... ,y(n)) = g(t); for example, y'' + 4y' + 4y = t.
    The solution to the homogeneous problem is a linear combination of e-2t and te-2t. As it turns out for my example, the particular solution is yp = -5/4 + t/4.

    We know that yc is a solution to the homogeneous problem, which means for my example, yc'' + 4yc' + 4yc = 0. We also know that yp is a particular solution of the nonhomogeneous problem, so yp'' + 4yp' + 4yp = t.

    Then for a general solution y = yc + yp, we will have y'' + 4y' + 4y = t, regardless of which linear combination of e-2t and te-2t we choose.
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