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DE question

  1. Nov 7, 2013 #1
    Hello - I asked a similar question before, but it was not resolved for me, and the person who answered was rude, so I did not continue the conversation.

    I read this here: http://tutorial.math.lamar.edu/Classes/DE/SecondOrderConcepts.aspx

    "If y_1(t) and y_2(t)are two solutions to a linear, homogeneous differential equation then so is
    y(t) = c_1 y_1(t) + c_2 y_2(t), and it states that this is the general solution.

    I don't understand this: if y_1(t) and y_2(t) are solutions, then we should be done, right? We have our solutions. Why are we interested in making another solution? And, why is the sum of the solutions with multipliers the "general solution", and the other two solutions, y_1(t) and y_2(t), not general?

    Thank you
  2. jcsd
  3. Nov 7, 2013 #2


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    Homework Helper

    It's not enough for a solution to satisfy the ODE; it must also satisfy the initial conditions, and for a second-order ODE there are two such conditions.


    The functions [itex]y_1(x) = \cos x[/itex] and [itex]y_2(x) = \sin x[/itex] are solutions of
    y'' = -y.

    Neither [itex]y_1[/itex] nor [itex]y_2[/itex] satisfy the intitial conditions [itex]y(0) = y'(0) = 1[/itex], but the linear combination [itex]y(x) = \cos x + \sin x = y_1(x) + y_2(x)[/itex] does, and in general [itex]a\cos x + b\sin x[/itex] is the solution of [itex]y'' = -y[/itex] which satisfies the initial conditions [itex]y(0) = a[/itex] and [itex]y'(0) = b[/itex].
  4. Nov 7, 2013 #3
    Thank you
  5. Nov 11, 2013 #4


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    Science Advisor

    The basic theorem here is that "the set of all solutions to a linear homogeneous differential equation of order n form an n dimensional vector space".

    In particular, that means there exist a "basis" for the vector space (solution set) consisting of n functions such that every solution can be written as a linear combination of those solutions.

    In the case of a "linear homogeneous second order equation", there must exist two independent solutions, [itex]y_1(x)[/itex] and [itex]y_2(x)[/itex] such that any solution, y(x), can be written [itex]y(x)= Ay_1(x)+ By_2(x)[/itex] for appropriate constants A and B.

    We are not "done" when we find [itex]y_1[/itex] and [itex]y_2[/itex] because there exist an infinite number of solutions.
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