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    Does a 3rd order differential equation whose auxiliary equation has one real and two complex solutions,

    ie. n = n_1, n = n_2 \pm n_3 i

    have the following complementary function?

    y = k_1 e^{n_1 x} + e^{n_2 x}(k_2 sin(n_3 x) + k_3 cos(n_3 x))
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    (Original post by Ateo)
    Does a 3rd order differential equation whose auxiliary equation has one real and two complex solutions,

    ie. n = n_1, n = n_2 \pm n_3 i

    have the following complementary function?

    y = k_1 e^{n_1 x} + e^{n_2 x}(k_2 sin(n_3 x) + k_3 cos(n_3 x))
    I would imagine so, although personally I've never come across a 3rd order DE in a real situation! The basic principle is that the CF will be a linear combination of functions e^{ax} where a is a root of the auxiliary equation.
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    (Original post by davros)
    I would imagine so, although personally I've never come across a 3rd order DE in a real situation! The basic principle is that the CF will be a linear combination of functions e^{ax} where a is a root of the auxiliary equation.
    I did manage to get that for real roots. However I was curious about the situation described above. The way I laid out the example above with the trigonometric functions is the way my book treats complex solutions for 2nd order DE's. However, I tried checking whether what I described above works with an example DE on wolfram alpha and the result had no trigonometric functions.
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    (Original post by davros)
    I would imagine so, although personally I've never come across a 3rd order DE in a real situation!
    Non-Thermal optical emission in extragalactic jets.
    See no. 19. http://iopscience.iop.org/0004-637X/499/1/167/fulltext/
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    (Original post by Ateo)
    I did manage to get that for real roots. However I was curious about the situation described above. The way I laid out the example above with the trigonometric functions is the way my book treats complex solutions for 2nd order DE's. However, I tried checking whether what I described above works with an example DE on wolfram alpha and the result had no trigonometric functions.
    I'm too lazy to make up an example, but I'm sure that if the DE (and hence the auxiliary eq) has real coefficients, then any complex roots occur in complex conjugate pairs so you get the type of answer you suggested
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    (Original post by davros)
    I'm too lazy to make up an example, but I'm sure that if the DE (and hence the auxiliary eq) has real coefficients, then any complex roots occur in complex conjugate pairs so you get the type of answer you suggested
    The thing is that I didn't with the example I put into wolfram alpha.
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    (Original post by Ateo)
    The thing is that I didn't with the example I put into wolfram alpha.
    Ah, right - you've picked an unfortunate case there () - one of the roots of the AE is m=0 so there are only 2 linearly independent exponential solutions and you have to introduce an extra term of the form kxe^{ax} - a bit like how you do when you're trying to find a particular integral and it matches the CF
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    (Original post by davros)
    Ah, right - you've picked an unfortunate case there () - one of the roots of the AE is m=0 so there are only 2 linearly independent exponential solutions and you have to introduce an extra term of the form kxe^{ax} - a bit like how you do when you're trying to find a particular integral and it matches the CF
    Oops, I made a typo, the DE was supposed to be:

    y^{(3)} - 2y'' + y' - 2y = 0

    In which case it does work out as expected. Thus my initial assumption seems correct.
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    (Original post by Ateo)
    Oops, I made a typo, the DE was supposed to be:

    y^{(3)} - 2y'' + y' - 2y = 0

    In which case it does work out as expected. Thus my initial assumption seems correct.
    Glad you spotted it - saved me doing any serious thinking
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    (Original post by davros)
    Glad you spotted it - saved me doing any serious thinking
    Well is there any harm in that ?
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    (Original post by Ateo)
    Well is there any harm in that ?
    There is at my age
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    (Original post by davros)
    There is at my age
    Does the enthusiasm and the so called 'freshness of mind' for mathematics really diminish that much with age? Or is it just a slight hint of sarcasm on your side?
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    (Original post by Ateo)
    Does the enthusiasm and the so called 'freshness of mind' for mathematics really diminish that much with age? Or is it just a slight hint of sarcasm on your side?
    It's more a case of competing priorities and other distractions in my case!!

    My mind is probably less agile than it used to be, and less able to focus on problems for an extended period of time, although I do think I have a much better perspective on maths these days - I never really thought I was seeing the bigger picture before
 
 
 
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