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ODE (solve for particular integral) am i right?

  1. Feb 24, 2009 #1
    ODE (solve for particular integral) am i right??

    1. The problem statement, all variables and given/known data
    That is all the examples that i know to determine the particular solution.
    are all of them correct? Note: e= exponent

    a)x^3 ==> Ax^3 + Bx^2 + Cx + D
    b)xe^2 ==> e^2(Ax +B)
    c)sin x ==> Asin x + Bcos x
    d)cos x ==> Asin x + Bcos x
    e)sin x + cos x ==> Asin x + Bcos x + Csin x + Dcos x
    f)sin x + xcos x ==> Asin x + Bcos x + (Cx+D)sin x + (Ex+F)cos x
    g)x^2(sin x) ==> (Ax+B)sin x + (Cx+D)cos x
    h)e^x (sin x) ==> e^x( Asin x + Bcos x)
    2. Relevant equations
    Am i right?
    is it still got another type of particular?
    that is all i know only.



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 24, 2009 #2
    Re: ODE (solve for particular integral) am i right??

    that looks about right, except for e) and g). also, on b, it it supposed to be xe^(2x)?

    on e), if you factor your answer you get
    Asin x + Bcos x + Csin x + Dcos x = sinx(A+C) + cosx(B+D). since A, B, C, D are just constants, you can tell that you don't realy need the C or the D.

    on g), you're missing an x^2 term.
     
  4. Feb 24, 2009 #3

    HallsofIvy

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    Re: ODE (solve for particular integral) am i right??

    One caveat: If a function of that form is already a solution to the associated homogeneous equation, then you will need to multiply by 4.

    For example, if the problem is y"- 5y'+ 6y= ex, so the characteristic equation is r2- 5r+ 6= (r- 3)(r- 2)= 0, which has roots 2 and 3, so that the solutions to the homogeneous equation are e2x and e3x, then, yes, I would try a particular solution of the form y= Aex.

    But if the problem is y"- 4y'+ 3y= ex, so the characteristic equationj is r2- 4r+ 3= (r-1)(r- 3)= 0, which has roots 1 and 3, so that the solutions to the homogeneous equation are ex and e3x, then Aex will give 0, not ex, for any A. Now I would try y= Axex.

    If the problem is y"- 2y'+ y= ex, so that the characteristic equation is r2- 2r+ 1= (r-1)2= 0, which has 1 as double root, so that the solutions to the homogeneous equation are ex and xex, then I would have to tr Ax2ex.
     
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