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Homework Help: Ambiguity in the method applied for differential equations

  1. Aug 5, 2016 #1
    1. The problem statement, all variables and given/known data
    Why do we need two solutions to solve a 2nd order linear differential equation?
    lets consider a differential equation with equal roots for auxiliary equation. So the reasoning behind why cant we use y=Aen1x+Ben2x
    as its general solution is because since the roots are equal we get a single solution y=Cenx
    . So my question is what's the problem with having one solution? After all it's A SOLUTION.
  2. jcsd
  3. Aug 5, 2016 #2
    Are you asking why, mathematically, do second order ODEs have two solutions, or why in general would we want to try to get a second solution if we already have one?

    The answer to the latter is: what if, given our initial conditions, the one solution we did find doesn't agree with what we find in the real world? We want to find the most general solution possible for it to actually match up with experiment. It is possible that one solution turns out to be always ##0## and that the other solution is the important one.
  4. Aug 5, 2016 #3
    Oh thank you for answering the latter. Btw I am asking both questions
  5. Aug 5, 2016 #4
    Thank you for explaining the second one. By the way I was asking both the questions
  6. Aug 5, 2016 #5


    Staff: Mentor

    It helps to think of the connection between differential equations and linear algebra, with one connection being that the solution space for a given differential equation is a function space, a type of vector space. The solution space for a first-order DE is of dimension one, the solution space for a second-order DE is of dimension two, and in general, the solution space for an n-th order DE is of dimension n.

    If a vector space or solution space is one-dimensional, every vector or function in the space can be written as a constant multiple of a single vector/function -- a basis vector/function. If a vector space or function space is of dimension two, any basis for the space must consist of two linearly independent vectors/functions. Every vector/function in the space can be written as a linear combination of the basis vectors/functions. That is, a sum of constant multiples of the two vectors/functions.

    In your example, the solution space is two dimensional, so to describe all solutions, you have to come up with a basis for the space -- two linearly independent functions. In this case, the functions ##e^{nx}## and ##xe^{nx}## will work as the basis.
  7. Aug 5, 2016 #6

    Ray Vickson

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    Homework Helper

    Generally (not always) there are two possible values for ##n##, so which one would you use? In some cases the root ##n## is a double root, so again there are two possible solutions ##e^{nx} ## and ##x e^{nx}##. Again, which one would you choose?

    In both the cases above you may need to take a linear combination of both solutions in order to fit additional conditions in the problem. Generally---in applications at least---differential equations do not come without some type of extra conditions, such as initial and/or final values, or values and derivatives specified at some point. When you have those, a single one of the two solutions may not solve the problem.
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