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Homework Help: Diff. Eq. Problem -> y'' - 2y' - 3y = -3 t e^(-t) What is general solution?

  1. May 26, 2005 #1
    Diff. Eq. Problem ---> y'' - 2y' - 3y = -3 t e^(-t) What is general solution?

    This is problem number 3 in section 3.6 in Boyce, DiPrima 8th Edition "Elementary Differential Equations and Boundry Value Problems"

    Find the general solution of the differential equation:

    [tex]y'' - 2y' - 3y = -3 t e^{-t}[/tex]

    Here is what I have done so far:

    The characteristic equation is [tex]r^2 - 2r - 3 = 0[/tex] and [tex]r_1 = 3, r_2 = -1[/tex]

    That gives [tex]y(t) = C_1 e^{3t} + C_2 e^{-t}[/tex] as the complementary solution.

    Now I run into trouble when selecting the superposition equation for using the method of undetermined coefficients. I tried [tex]Y(t) = (A t + B) e^{-t}[/tex] with no luck. I get [tex]A = \frac{3}{4} t[/tex] for an answer, which is incorrect.

    The book lists the answer as [tex]y(t) = C_1 e^{3t} + C_2 e^{-t} + \frac{3}{16} t e^{-t} + \frac{3}{8} t^2 e^{-t}[/tex]

    What am I doing wrong? Is the [tex]Y(t)[/tex] I chose correct? Please help!

    Thank you.
     
  2. jcsd
  3. May 26, 2005 #2

    arildno

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    Try a particular solution [tex]Y(t)=(At^{2}+Bt)e^{-t}[/tex]
    and determine A, B.
     
  4. May 26, 2005 #3
    Thanks

    I tried it and it sounds more reasonable. Why must this be the particular solution instead of the one I origianlly chose?

    I'm still having trouble. I have the Y, Y', Y'' all substituted and I get [tex]-6A - 4B = -3t[/tex] How do I solve this for A and B? There must be a second equation in there somewhere right?
     
  5. May 26, 2005 #4

    HallsofIvy

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    The reason (At+B)e-t didn't work is that e-t is already a solution to the homogeneous differential equation. You should have learned that, in situations like that, you multiply by t (or higher powers of t is te-t, etc. are also solutions). Here you should try, as Arildno said, t(At+ B)e-t= (At2+ Bt)e-t.
    Now you say you got -6A- 4B= -3t. How?

    If y= (At2+ Bt)e-t, then y'= -(At2+ Bt)e-t+ (2At+ B)e-t and y"= (At2+ Bt)e-t- 2(2At+B)e-t+ 2Ae-t

    y"= (At2+ Bt)e-t- 2(2At+B)e-t+ 2Ae-t
    -2y'= 2(At2+ Bt)e-t-2(2At+ B)e-t
    -3y= -3((At2+ Bt)e-t)

    The first terms cancel leaving -4(2At+B)e-t+2Ae-t= -3te-t. Dividing both sides by e-t leaves
    -4(2At+ B)+ 2A= -3t or -8At+ (-4B+2A)= -3t. We must have -8A= -3 and -4B+2A= 0.
     
  6. May 26, 2005 #5

    dextercioby

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    Try the method of Lagrange of varying the constants.

    Daniel.
     
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