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Math Help - 2nd Order Linear O.D.E. Help

  1. #1
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    2nd Order Linear O.D.E. Help

    Hi guys, I've got this differential equation:

    \ddot{y}+y=4sin{\alpha{x}}

    And I've worked out the general solution when it is set equal to zero to be:

    Acos{x}+Bsin{x}

    I need help finding the equation to set y(x) to so that i can solve the rest of it. Any clues?

    Thanks in advance
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    Senior Member Peritus's Avatar
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jwade456 View Post
    Hi guys, I've got this differential equation:

    \ddot{y}+y=4sin{\alpha{x}}

    And I've worked out the general solution when it is set equal to zero to be:

    Acos{x}+Bsin{x}

    I need help finding the equation to set y(x) to so that i can solve the rest of it. Any clues?

    Thanks in advance
    no, your particular solution must be of the form A \sin \alpha x + B \cos \alpha x
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    Quote Originally Posted by Jhevon View Post
    no, your particular solution must be of the form A \sin \alpha x + B \cos \alpha x
    Unless \alpha = 1 ........
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Unless \alpha = 1 ........
    it should work even if \alpha = 1, shouldn't it?
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    it should work even if \alpha = 1, shouldn't it?
    Have you tried it ......
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Have you tried it ......
    haha, no. but that wink makes me want to take your word for it
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    Quote Originally Posted by Jhevon View Post
    haha, no. but that wink makes me want to take your word for it
    lol! Quite wise. When \alpha = 1 the proposed form of particular solution is actually the homogenous solution, that is, the solution to \ddot{y}+y = 0.
    Last edited by mr fantastic; October 17th 2008 at 07:17 PM. Reason: Corrected a careless copy and paste
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    So when  \alpha = 1 what form would you have to take to solve the equation?

    Furthermore, do you have to solve the equation with two different forms depending on whether or not alpha equals one, or is there a form that will work regardless of alpha's value?

    Cheers.
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    Quote Originally Posted by U-God View Post
    So when  \alpha = 1 what form would you have to take to solve the equation?

    [snip]
    When \alpha = 1, use the form (x + Cx^2) (A \cos x + B \sin x) to construct a particular solution.

    Quote Originally Posted by U-God View Post
    [snip]
    Furthermore, do you have to solve the equation with two different forms depending on whether or not alpha equals one,
    [snip]
    Yes.

    Quote Originally Posted by U-God View Post
    [snip]
    or is there a form that will work regardless of alpha's value?

    Cheers.
    No.
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    Thanks mr fantastic, you certainly live up to your name!!

    If I may ask just one more question, where did you get the form:  (x + Cx^2) (A \cos x + B \sin x) from?

    Is there a way I can derive it myself? Or is it just a rule of thumb.

    Thanks again!
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    Quote Originally Posted by U-God View Post
    Thanks mr fantastic, you certainly live up to your name!!

    If I may ask just one more question, where did you get the form:  (x + Cx^2) (A \cos x + B \sin x) from?

    Is there a way I can derive it myself? Or is it just a rule of thumb.

    Thanks again!
    The informal answer is that it's the simplest form that works .....

    (A \cos x + B \sin x) obviously is no good so try the next simplest: x (A \cos x + B \sin x).

    That doesn't work so try the next simplest: (x + C x^2) (A \cos x + B \sin x). That works so it's all you need. Consider it a rule of thumb.

    There is a way of getting it formally but I don't have the time to type out the details right now. Chris likes D.E.'s ... he might get a kick out of doing it.
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    You have helped soooo much. Thanks again.
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Chris likes D.E.'s ... he might get a kick out of doing it.
    Haha

    But like you, I don't have the time right now...

    I'll take a look at it later.

    --Chris
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    Quote Originally Posted by mr fantastic View Post
    The informal answer is that it's the simplest form that works .....

    (A \cos x + B \sin x) obviously is no good so try the next simplest: x (A \cos x + B \sin x).

    That doesn't work so try the next simplest: (x + C x^2) (A \cos x + B \sin x). That works so it's all you need. Consider it a rule of thumb.

    There is a way of getting it formally but I don't have the time to type out the details right now. Chris likes D.E.'s ... he might get a kick out of doing it.
    Thanks again,

    I handed my work in and it all looks good, however some of my friends said that they did it by using the form  x (A \cos x + B \sin x) .
    You said here that that form wouldn't work and to move on to the next form up, so out of interest when I got home I tried it with form:  x (A \cos x + B \sin x) and lo and behold I got the same answer as with form: (x + C x^2) (A \cos x + B \sin x).

    Does that mean I did something wrong?
    Cheers,
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