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Math Help - initial value problem

  1. #1
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    initial value problem

    Consider the initial value problem x''+x+\varepsilon x=0 with x(0)=1, x'(0)=0. I need to find the exact solution to this problem. Thanks.
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    Quote Originally Posted by splash View Post
    Consider the initial value problem x''+x+\varepsilon x=0 with x(0)=1, x'(0)=0. I need to find the exact solution to this problem. Thanks.
    I assume \epsilon is constant? Then
    x'' + (1 + \epsilon)x = 0

    The characteristic equation is m^2 + (1 + \epsilon) = 0, so
    m = \pm i \sqrt{1 + \epsilon}. So the general solution is:

    x(t) = Ae^{it\sqrt{1 + \epsilon}} + Be^{-it\sqrt{1 + \epsilon}}

    A physicist would probably prefer the equivalent expression:
    x(t) = Csin(t\sqrt{1 + \epsilon} ) + Dcos(t \sqrt{1 + \epsilon})

    Now use the form of your choice and apply the initial conditions.

    -Dan
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    How did you get the characteristic equation? The way I remember is for matricies.
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    Quote Originally Posted by splash View Post
    How did you get the characteristic equation? The way I remember is for matricies.
    I was not aware that the "characteristic equation" could be used in that regard. I thought it was solely to describe the solutions of:

    Characteristic equation (of A):

    det(A - LI) = 0, where L is lambda;

    or, similarly, det(LI - A) = 0

    If not, I learned something new.
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    Quote Originally Posted by AfterShock View Post
    I was not aware that the "characteristic equation" could be used in that regard. I thought it was solely to describe the solutions of:

    Characteristic equation (of A):

    det(A - LI) = 0, where L is lambda;

    or, similarly, det(LI - A) = 0

    If not, I learned something new.
    Isn't A a matrix?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by splash View Post
    How did you get the characteristic equation? The way I remember is for matricies.
    Quote Originally Posted by AfterShock View Post
    I was not aware that the "characteristic equation" could be used in that regard. I thought it was solely to describe the solutions of:

    Characteristic equation (of A):

    det(A - LI) = 0, where L is lambda;

    or, similarly, det(LI - A) = 0

    If not, I learned something new.

    Perhaps I have the wrong term. I was too lazy to look it up.

    -Dan

    Edit: Ha! My ODE book does call it a characteristic equation, but you might know it better by the term "auxiliary equation" perhaps.
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    Quote Originally Posted by splash View Post
    Isn't A a matrix?
    Yes, A is a matrix. I meant to quote topsquark and not you.
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    Quote Originally Posted by topsquark View Post
    Perhaps I have the wrong term. I was too lazy to look it up.

    -Dan
    Oh ok. Then how did you get the m^2 + (1 + \epsilon) = 0 equation. Sorry to pester, I'm just confused about where that came from. Thanks.
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by splash View Post
    Oh ok. Then how did you get the m^2 + (1 + \epsilon) = 0 equation. Sorry to pester, I'm just confused about where that came from. Thanks.
    Given the ODE a_ny^{(n)} + a_{n-1}y^{(n-1)} + ... + a_1y' + a_0 = 0 (with constant coefficients) you may construct a general solution by solving the polynomial equation:
    a_nm^n + a_{n-1}m^{n-1} + ... + a_1m + a_0 = 0
    (the characteristic or auxiliary equation)

    Solutions to the ODE will be:
    y = Ae^{m_1x} + Be^{m_2x} + ... + Ze^{m_nx}
    where the \{ m_i \} are the solutions to the polynomial and A, B, ...Z are constants of integration.

    -Dan
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    Thanks a lot tops. Very helpful.
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    Quote Originally Posted by topsquark View Post
    I assume \epsilon is constant? Then
    x'' + (1 + \epsilon)x = 0

    The characteristic equation is m^2 + (1 + \epsilon) = 0, so
    m = \pm i \sqrt{1 + \epsilon}. So the general solution is:

    x(t) = Ae^{it\sqrt{1 + \epsilon}} + Be^{-it\sqrt{1 + \epsilon}}

    A physicist would probably prefer the equivalent expression:
    x(t) = Csin(t\sqrt{1 + \epsilon} ) + Dcos(t \sqrt{1 + \epsilon})

    Now use the form of your choice and apply the initial conditions.

    -Dan
    How do you know that,
    1+\epsilon<0
    And if it is, then it should be,
    \pm i \sqrt{-(1+\epsilon)}
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  12. #12
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    How do you know that,
    1+\epsilon<0
    And if it is, then it should be,
    \pm i \sqrt{-(1+\epsilon)}
    Good point! (As usual. )

    -Dan
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