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Is our complementary solution guaranteed to be linearly independent?

  1. Sep 5, 2012 #1
    I had kind of a general question. Say I have a second order, homogeneous ODE. Say I use one of the general techniques to generate a complementary solution for my ODE and I end up with something of the form y = C1(solution1) + C2(solution2)

    Am I gauranteed that these two solutions will be linearly independent or do I need to verify with the Wronskian each time?

    Also, can I think of linearly independent solutions as a "basis" like I do in linear algebra? It seems like it because its my understanding that any solution for the ODE is simply a linear combination of solutions in this "basis"

    Also, can we tell the dimensions of this basis based on the order of the ODE? Like, a second order ODE has two solutions in its basis, a third order has 3, a first order has one, etc.

  2. jcsd
  3. Sep 5, 2012 #2
    Yes, you can think of the solutions as a basis in a vector space. In this case, they're basis functions in a space of functions, but most of the ideas from vector algebra still apply.

    I think often the techniques chosen to generate solutions are meant to come up with solutions out of orthogonal functions--sines and cosines, for instance, are automatically linlearly independent as long as none of their frequencies are the same. Someone may know more on this matter, though.
  4. Sep 5, 2012 #3


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    A complementary solution can't be "dependent" on the solution to the homogeneous equation. Suppose the equation is L(f)= g(x) where L is the linear differential operator, u(x) a linear combination of the independent solutions to the homogenelus equation, and v(x) a complementary solution. To be dependent means that there exist numbers, a and b, non-zero, such that au(x)+ bv(x)= 0. Applying L to both sides of that, L(au(x)+ bv(x))= aL(u)+ bL(v)= a(0)+ b(g)= 0 so that bg(x)= 0 for all x. Since g is not always 0, b= 0. That makes our original equation au(x)= 0 for all x and so a= 0.

    The solutions to any linear homogeneus nth order differential equation form a vector space of dimension n. The solutions to a non-homogeneous linear equation do NOT form a vector space but do form a "linear manifold". If you think of a one-dimensional vector space as a line through the origin of a xy-coordinate system, or a two dimensional vector space as a plane through the origin of a xyz-coordinate system, then you can think of a one dimensional "linear manifold" as a line NOT through the origin or a two dimensional "linear manifold" NOT through the origin.
    Last edited by a moderator: Sep 5, 2012
  5. Sep 5, 2012 #4

    Thank you, this really does help me visualize this.

    So I want to make sure I understand. So for example, when dealing with non homogeneous equations, does it mean we can't establish a "basis" or fundamental set of solutions so that every particular solution can be expressed as a linear combination of those solutions?

    THe thing that prompted me to ask was this particular problem:

    y'' - 5y' + 4y = 8ex

    I can generate a basis for the vector space of solutions for the associated homogeneous equation and its B = {e4x,ex} (I got this by solving the auxilary equation and attaining distinct real roots x=4 and x=1)

    Using variation of parameters, I can then generate a particular solution for this sytem:


    To make sure it is indeed a particular solution I took y'p and y''p and plugged it in to the original ODE and it does check out. However, yp is not a linear combination of the functions in B.
  6. Sep 5, 2012 #5
    oh wait I think I get it, the "basis" as I'm thinking of it for a nohomogeneous linear 2nd order ODE would be B={b1,b2,yp}

    where b1 and b2 are the functions in the basis for the associated homogeneous ODE and yp is a particular solution satisfying the non homogeneous ODE.

    EDIT: Oh waaaait... No what I just wrote is wrong. Any solution Y(x) of the non homogeneous system can be expressed as:


    So I can see why {b1,b2,yp} is NOT a "basis" because we can't just have any linear combination of these to generate solutions, the solutions are linear combinations of b1 and b2 with yp simply added on (not any multiple of yp)! So this is why it's not a vector space but a linear manifold as you described it
    Last edited: Sep 5, 2012
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