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Math Help - Second Order Differential Equations.

  1. #1
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    Second Order Differential Equations.

    Can someone give me a convincing explanation as to why when solving a second differential equation:
    e.g y" + 3y'-4y = 3x+5

    That you must ADD the CF to the PI?
    - I know it works but i want to know why? I don't want to just demonstrate it either.

    It doesn't make sense to me to just say to complete the solution we need to solve y" + 3y'-4y= 0 . Why zero?

    The PI clearly works. But also just because we are after some constants doesn't necessarily mean we should put y" + 3y'-4y=0?
    Last edited by mr fantastic; February 1st 2011 at 01:43 AM. Reason: Deleted begging from title.
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  2. #2
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    I'll give you one good reason: a particular integral has no arbitrary constants. Therefore, you cannot play around with the PI to get it to fit initial or boundary conditions. The complementary function does have arbitrary constants. Hence, you get CF + PI to satisfy initial or boundary conditions.
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  3. #3
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    I am not sure i buy that. If we are just after arbitrary constants we can pick any function but why choose the CF? and why must the DE=0?
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  4. #4
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    You can't pick any ol' function, because you want your result to satisfy the DE. The CF has two properties that make it ideal for satisfying initial conditions: 1. It has two arbitrary constants. 2. It satisfies the homogeneous DE, which means I can change those constants all day long without changing the fact that the CF + PI still satisfies the DE. If you do that with any other function, the result will not satisfy the DE.
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  5. #5
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    For the simple reason that you have to recognise that the RHS is actually \displaystyle 0 + 3x + 5, which means to have the most general solution, you need a solution that satisfies the \displaystyle 0 as well as the \displaystyle 3x+5.
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  6. #6
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    BUT why can you do this now but not in the case below?

    Because lets look at a quadratic:

    x^2 + 5x +4 =2

    You can't say the solution consists of x^2+5x +4 =0
    and x^2+ 5x+4 =2

    So why can you do this now?
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    Because your second-order ODE operator is still linear, though your quadratic equation is not linear. If Ly is the LHS of your ODE, where L represents the differential operator

    L=D^{2} + 3D-4,

    then you can convince yourself that if y_{1} and y_{2} are both solutions to the homogeneous ODE Ly=0, then any linear combination ay_{1}+by_{2} is also a solution to Ly=0.

    This fact is not true of your quadratic.
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  8. #8
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    Thanks, that is starting to make sense.. Can you give me another example of a linear operator? A simple one?
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    Quote Originally Posted by Ackbeet View Post
    Because your second-order ODE operator is still linear, though your quadratic equation is not linear. If Ly is the LHS of your ODE, where L represents the differential operator

    L=D^{2} + 3D-4,


    then you can convince yourself that if y_{1} and y_{2} are both solutions to the homogeneous ODE Ly=0, then any linear combination ay_{1}+by_{2} is also a solution to Ly=0.

    This fact is not true of your quadratic.
    And how does the PI fit into this notation?
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  10. #10
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    Sure. Here's a linear operator:

    A=\begin{bmatrix}0&1\\ 1&0\end{bmatrix},

    which has the effect of taking points (x,y) in the xy plane and reversing their coordinates. That is, (x,y)\overset{A}{\mapsto}(y,x). Example:

    \begin{bmatrix}2\\3\end{bmatrix}=\begin{bmatrix}0&  1\\ 1&0\end{bmatrix}\begin{bmatrix}3\\2\end{bmatrix}  .

    If you want another simple differential operator, try this:

    L=D-1.

    Here's a nonlinear operator:

    L=(1-|y|)D.

    On to your next post: the PI fits into the notation scheme as follows: the particular integral y_{p} satisfies the following equation:

    Ly_{p}=3x+5, going back to your original DE.
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  11. #11
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    Quote Originally Posted by Ackbeet View Post
    Because your second-order ODE operator is still linear, though your quadratic equation is not linear. If Ly is the LHS of your ODE, where L represents the differential operator

    L=D^{2} + 3D-4,

    then you can convince yourself that if y_{1} and y_{2} are both solutions to the homogeneous ODE Ly=0, then any linear combination ay_{1}+by_{2} is also a solution to Ly=0.

    This fact is not true of your quadratic.
    Ackbeet,

    Are you saying then that there is no solution in the case of the quadratic ?

    Thanks
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  12. #12
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    Reply to bugatti79,

    No, I'm not saying there's no solution to the quadratic equation. What I am saying is that, in the case of the quadratic, you can't expect to add up linear combinations of solutions to the quadratic, and end up with another solution to the quadratic. You can do that with the solutions to a linear homogeneous ODE (or PDE, for that matter).
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