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Second Order Nonhomogenous Differential Equation

  1. Nov 1, 2013 #1
    Hello everyone, I'm having trouble understanding the solutions to DE's of the form:

    [tex]ay''+by'+cy=f(t)[/tex]

    We've gone over them in class, I've talked with my friends, and it just doesn't make any sense to me. I was wondering if anyone on here would help me understand the solutions, it would be much appreciated.
     
  2. jcsd
  3. Nov 1, 2013 #2

    vanhees71

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    It's a linear equation (supposed that [itex]a, \quad b, \quad c[/itex] are functions of the independent variable [itex]x[/itex] but not of the unknown function [itex]y[/itex]), which makes the task to solve it easier.

    First of all, suppose you have two solutions [itex]y_1(x)[/itex] and [itex]y_2(x)[/itex]. Now consider [itex]y(x)=y_1(x)-y_2(x)[/itex]. Now, because differentiation wrt. [itex]x[/itex] is a linear operation we have
    [tex]a y''+b y' + c y = (a y_1''+by_1'+c y_1)-(a y_2''+b y_2' +c y_2)=f-f=0.[/tex]
    This means that the difference of two solutions of the inhomogeneous equations is always a solution of the homogeneous equations.

    In turn we can conclude that any solution of the inhomogeneous equation is given as the sum of one particular solution of the inhomogeneous equation [itex]y_p(x)[/itex] and the general solution of the homogeneous equation [itex]y_h(x)[/itex].

    Further it's clear that with any set of solutions of the homogeneous equation also any linear combination of such solutions is again a solution of the homogeneous equation. The solutions of the homogeneous equation thus build a linear subspace in the vector space of twice differentiable functions. One can prove that this subspace is two-dimensional (look for "Wronskian Determinant" in your textbook or online), i.e., you need to find two linearly independent solutions of the homogeneous equation [itex]y_{h1}(x)[/itex] and [itex]y_{h2}(x)[/itex]. Linearly independent means simply that [itex]y_{h1}(x)/y_{h2}(x) \neq \text{const}[/itex]. Then the general solution of the inhomogeneous system reads
    [tex]y(t)=C_1 y_{h1}(x) + C_2 y_{h2}(x)+y_p(x).[/tex]
    Here, [itex]C_1[/itex] and [itex]C_2[/itex] are arbitrary constants.

    A solution can be uniquely determined by imposing, e.g., initial conditions,
    [tex]y(t_0)=y_0, \quad y'(t_0)=y_0'.[/tex]
     
  4. Nov 2, 2013 #3
    Thank you for spelling this out for me. It already is starting to make more sense. I'll have to spend the next couple of days studying your reply though, as I am busy all weekend. :)
     
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