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Homework Help: Second order DE from my midtearm

  1. Jun 26, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider the ODE:

    y''+p(x)y'+q(x)y=g(x)

    It is given that the functions y=x[itex]^{2}[/itex], y=x and y[itex]\equiv[/itex]1
    are solutions of the equation.
    Find the general solution of the equation.

    2. Relevant equations

    3. The attempt at a solution

    Well, given the three solutions, and taking into account that they are mutually linearly independent, I would say that any couple of these functions will give me a set that forms the general solution.

    If i take y=x and y[itex]\equiv[/itex]1, I get the general solution:

    y(x)=c1+c2*x

    Now, I know that x[itex]^{2}[/itex] is also a solution, but it can't be derived from the general solution, so obviously I'm doing something wrong.

    I also thought of the possibility that y=1 is a singular solution and the general solution is:

    y(x)=c1*x+c2*x^2

    does this makes sense?

    This is a question from my midterm and i still can't figure it out. So... help?
     
  2. jcsd
  3. Jun 26, 2011 #2

    tiny-tim

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    Welcome to PF!

    Hi Danny B! Welcome to PF! :smile:

    (try using the X2 and X2 icons just above the Reply box :wink:)

    Each of those three are particular solutions …

    Hint: what relation is there between any two particular solutions (ie any two solutions)? :wink:
     
  4. Jun 27, 2011 #3
    Thank you.

    I think I got it now.

    My mistake was to use the principle of superposition with a non-homogeneous equation.

    So the general solution should be:

    (Considering that any difference of the particular solutions is a homogeneous solution)

    y[itex]_{(x)}[/itex]=c[itex]_{1}[/itex](x-1)+c[itex]_{2}[/itex](x[itex]^{2}[/itex]-1)+Y[itex]_{p}[/itex]

    But how do I choose Y[itex]_{p}[/itex]?

    Can I just use any of the three particular solutions?
     
  5. Jun 27, 2011 #4

    tiny-tim

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    Hi Danny B! :smile:

    Let's write it out in full, so you can see how it works …

    the general solution can be written in three ways

    A(x2 - 1) + B(x - 1) + 1

    A(x2 - x) + x + B(1 - x)

    x2 + A(x - x2) + B(1 - x2)​

    the homogeneous general solution can be written in three ways

    A(x2 - 1) + B(x - 1)

    A(x2 - x) + B(1 - x)

    A(x - x2) + B(1 - x2)​

    now you need to spend ten minutes convincing yourself that they're the same! :rolleyes:

    he he :biggrin:
     
  6. Jun 27, 2011 #5
    Thanks a lot!

    You've been a great help!
     
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