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Differential Equations help

  1. Jul 27, 2007 #1
    1. The problem statement, all variables and given/known data

    1) [tex]y''' - y' = 2\sin{t}[/tex], find the general solution.

    2) [tex]x^2y'' + 3xy' + 5y = 0[/tex], find the general solution.

    I'm pretty sure I did #2 correct, but I'm stuck on #1. I can't find the particular solution. I would, however, like to know if I did #2 right.

    2. Relevant equations

    [tex]F(r) = r^2 + (\alpha - 1)r + \beta = 0[/tex]

    For [tex]r_1 \ and \ r_2[/tex] Complex conjugates:

    [tex]y = C_1x^{\lambda}\cos(\mu\ln{x}) + C_2x^{\lambda}\cos(\mu\ln{x})[/tex]

    3. The attempt at a solution

    1) First I found the homogenous solution:

    [tex]y''' - y' = 0[/tex]

    [tex]r^3 - r = 0[/tex]

    [tex]r(r + 1)(r - 1) = 0[/tex]

    [tex]r_1 = 0, \ r_2 = -1, \ r_3 = 1[/tex]

    [tex]y = C_1 + C_2e^{-t} + C_3e^t[/tex]

    Now I try finding the particular solution by using the method of undetermined coefficients. Since [tex]y''' - y' = 2\sin{t}[/tex], I will assume [tex]y_1 = A_1tcost + A_2tsint[/tex], from this:

    [tex]y_1' = (A_1 + A_2t)\cos{t} + (A_2 - A_1t)\sin{t}[/tex]

    [tex]y_1'' = (2A_2 - A_1t)\cose{t} - (2A_1 + A_2t)\sin{t}[/tex]

    [tex]y_1''' = (A_1t - 3A_2)\sin{t} - (3A_1 + A_2t)\cos{t}[/tex]

    Now I plug that into the original equation, simplify and get:

    [tex](2A_1t - 4A_2)\sint{t} - 2(2A_1 + A_2t)\csot{t} = 2\sin{t}[/tex]

    This is where I'm stuck. How do I find [tex]A_1[/tex] and [tex]A_2[/tex]?

    2) [tex]x^2y'' + 3xy' + 5y = 0[/tex]

    [tex]\alpha = 3, \ \beta = 5[/tex]

    So:

    [tex]F(r) = r^2 + (\alpha - 1)r + \beta = 0[/tex]

    [tex]r^2 + 2r + 5 = 0[/tex]

    [tex]r_1 = -1 + 2i, \\ r_2 = 1 - 2i[/tex]

    So the general solution is:

    [tex]y = C_1x^{-1}\cos(2\ln{x}) + C_2x^{-1}\sin(2\ln{x})[/tex]

    Any help would be greatly appreciated.

    EDIT: I changed my assumption in #1. Since [tex]cos t[/tex] and [tex]sin t[/tex] are not solutions of the homogenous equation, I chose:

    [tex]y_1 = A_1\sin {t} + A_2\cos{t}[/tex]

    [tex]y_1' = A_1\cos{t} - A_2\sin {t}[/tex]

    [tex]y_1'' = -A_1\sin {t} - A_2\cos{t}[/tex]

    [tex]y_1''' = -A_1\cos{t} + A_2\sin {t}[/tex]

    Now I plug back in and get:

    [tex]-A_1\cos{t} + A_2\sin{t} - (A_1\cos{t} - A_2\sin {t}) = 2\sin{t}[/tex]

    [tex]2A_2\sin {t} - 2A_1\cos{t} = 2\sin {t}[/tex]

    [tex]A_2 - A_1\cot{t} = 1[/tex]

    So now I get [tex]A_1 = 0[/tex] and [tex]A_2 = 1[/tex]

    Therefore:

    [tex]y = C_1 + C_2e^{-t} + C_3e^t + \cos{t}[/tex][/tex]

    Is this correct?
     
    Last edited: Jul 27, 2007
  2. jcsd
  3. Jul 27, 2007 #2
    Have you tried using just A1*sin(t) +A2*cos(t)?
     
  4. Jul 27, 2007 #3
    Yes I just edited my first post to show that I did. Thanks.
     
  5. Jul 28, 2007 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Try differentiating y(t)= C1+ C2e-t+ C3et+ cos(t) three times and plug the result into the differential equation!
     
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