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Math Help - Second Order Differential Equation

  1. #1
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    Second Order Differential Equation

    Hi!

    How do I determine the particular integral for second order differential equations with mixed f(x)s? Like, it's neither strictly polynomial, trigonometry or exponential.

    Example of questions,

    (i) y'' - 4y' + 5y = (16x + 4)e^{3x}
    (ii) y'' + 3y' = (10x + 6)sin x
    (iii) y'' - 2y' + 4y = 541e^{2x}cos5x

    Thank you! (:
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  2. #2
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    Hey, each of those have right members which are particular solutions to some homogeneous differential equation. For example, the first one has a right member which is a solution to the equation:

    (D-3)^2 y=0

    So the operator (D-3)^2 becomes an "annihilation" operator that we can apply to both side of the equation to convert it to a homogeneous equation:

    (D-3)^2 (D^2-4D+5)y=0

    This is the method of undetermined coefficients. Are you familiar with that method? Try it first on some simple ones. Any DE book should have a section on this subject.
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  3. #3
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    Sorry but I do not understand what are you talking about at all.
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by shawsend View Post
    Hey, each of those have right members which are particular solutions to some homogeneous differential equation. For example, the first one has a right member which is a solution to the equation:

    (D-3)^2 y=0

    So the operator (D-3)^2 becomes an "annihilation" operator that we can apply to both side of the equation to convert it to a homogeneous equation:

    (D-3)^2 (D^2-4D+5)y=0

    This is the method of undetermined coefficients. Are you familiar with that method? Try it first on some simple ones. Any DE book should have a section on this subject.
    Aha!! I'm not the only one to know about the Annihilator approach

    Quote Originally Posted by panda* View Post
    Sorry but I do not understand what are you talking about at all.
    Read post #6 and #7 here to see how to tackle equations like these.

    --Chris
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  5. #5
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    Smile

    1. First you must solve the homogeneous equations
    (i) y''-4y-+5y=0
    (ii) y''+3y'=0
    (iii) y''-2y'+4y=0
    and you will find the general solutions for them.
    2. For non homogeneous equations you have to find the partial solutions according to what is on the right side of the equations
    (i)
    (ii)
    (iii)
    3. The final solution of your equations is the sum of the general solutions from 1st point and partial solutions from the 2nd point.
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  6. #6
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    Thank you for referrals of posts and helps.

    But I still don't really get it after I read the posts

    Generally, I do know how to handle second order differential equations if f(x) was one of the normal terms like, purely exponential, or trig or polynomials. But when it is multiplied together like that, I can't immediately identify the pattern of the particular integral.

    For example, the particular integral form for an exponential function of  f(x) = e^{kx} would be y_p = pe^{2x} and we differentiate and sub it back in to compare coefficients in order to obtain p.

    How do I deal when f(x) is not purely of those forms but multiplied together?

    Thank you.
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  7. #7
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    I'll work the first one using undetermined coefficients; differential equations open a unique window into the universe such that all her secrets are revealed:

    (D-3)^2(D^2-4D+5)y=0

    The general solution from the auxiliary equation is then:

    y=c_1 e^{3x}+c_2 e^{3x}+c_4 e^{2x}\cos(x)+c_5 e^{2x}\sin(x)

    with the desired solution (I'm taking this right out of Rainville almost word for word):

    y=y_c+y_p

    where y_c=c_4 e^{2x}\cos(x)+c_5 e^{2x}\sin(x). Then there must be a particular solution of the original equation containing at most the remaining terms: y_p=Ae^{3x}+Bxe^{3x}. That's the undetermined coefficients which can be determined by substituting this y_p into the original DE:

    When I do that I get:

    2Ae^{3x}+2Bxe^{3x}+2Be^{3x}=16xe^{3x}+4e^{3x}

    Equating coefficients, I get B=8 and A=-6. Then the general solution of y''-4y'+5y=(16x+4)e^{3x} is:

    y(x)=-6e^{3x}+8xe^{3x}+c_1e^{2x}\cos(x)+c_2e^{2x}\sin(x)
    Last edited by shawsend; November 17th 2008 at 03:09 AM.
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  8. #8
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    Although I still do not understand the method you recommended, but I'm still thankful for your generous help and time! I figured out a different approach to do it already But still thanks anyway!
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  9. #9
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    Quote Originally Posted by panda* View Post
    I figured out a different approach to do it already
    Close enough.
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