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Differential Equation

  1. May 17, 2014 #1
    1. The problem statement, all variables and given/known data

    I have a differential equation: [itex]\ddot{x} -2\dot{x} + 5x= 10 + 13cos(3t)[/itex]


    2. Relevant equations

    x(t) = xc + xp
    where xc is the Complementary Function and xp is the Particular Integral.


    3. The attempt at a solution

    I have formed and solved the auxiliary equation:

    [itex]m^{2} - 2m + 5 = 0[/itex]

    [itex]m_{1} = 1 + 2i , m_{2} = 1 - 2i[/itex]

    How would I go about forming xc = et(Acos2t + Bsin2t) ?

    Furthermore, how would I choose a trial function to determine the Particular Integral?
     
    Last edited: May 17, 2014
  2. jcsd
  3. May 17, 2014 #2

    LCKurtz

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    That's "complementary".

    I don't understand that question. You did just "form" it, didn't you?

    Well, since you need a constant and a ##\cos t## after you differentiate your ##x_p## I would suggest something of the form ##x_p = A + B\cos t + C\sin t##.
     
  4. May 17, 2014 #3
    I give up with English. Fixed the OP.

    We have taught to from what is called an auxiliary function by taking the coeffeicents of [itex]\ddot{x}, \dot{x}, [/itex]and [itex]x[/itex] and (say a, b, c) and forming the equation

    [itex]am^{2} + bm + c[/itex]

    and then solving this equation to find 2 values of m (m1 and m2) which are to then form the basis of the complementary function xc. The complementary function I've given above is what I have to form given my values m1 and m2.

    That makes sense, thanks. In the given answer a constant 3 has been added so that it reads:

    ##x_p = A + B\cos 3t + C\sin 3t##. Is there any particular reason that has been chosen?
     
  5. May 17, 2014 #4

    LCKurtz

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    Your original post had a ##\cos t## on the right side. Was that a typing error and it was supposed to be ##\cos(3t)##? That would be the only reason to use ##3t## in the trig functions for your ##x_p##.
     
  6. May 17, 2014 #5
    Ahh yes, my mistake. Thank you.

    Do you have any idea about forming the complementary function?
     
  7. May 17, 2014 #6

    Mark44

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    You already have the complementary function (i.e., the solution to the homogeneous problem). Are you asking about how to write a particular solution (not the complementary solution)? (BTW, the p subscript in xp stands for "particular.")

    If not, then I don't understand what you're having trouble with.
     
  8. May 17, 2014 #7
    I don't know how to go from having m1 and m2 to making the equation for xc. The equation in my post has been given to me in the answer, I don't know how to get it myself.
     
  9. May 17, 2014 #8

    LCKurtz

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    Your characteristic equation comes by looking for a solution ##x=e^{mt}## and you have found that ##m = 1\pm 2i##. That means the general solution of the homogeneous equation is$$
    x_c = Ae^{(1+2i)t} + Be^{(1-2i)t}= e^{t}(Ae^{2it} + Be^{-2it})$$Now use the Euler formulas for the complex exponentials in terms of sines and cosines to get it in the {sine,cosine} form.
     
  10. May 18, 2014 #9
    ##x_c = Ae^{(1+2i)t} + Be^{(1-2i)t}= e^{t}(Ae^{2it} + Be^{-2it})##

    I did this, but found that I get:

    ##x_c = e^{t}((A + B)cos2t + (A - B)isin2t)##

    Can these arbitrary constants be combined to yield something like:

    ##x_c = e^{t}(Ccos2t + Disin2t)## for some arbitrary constants C, D ?
     
    Last edited: May 18, 2014
  11. May 18, 2014 #10

    HallsofIvy

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    Yes, of course. Since A and B are arbitrary constants, so are A+ B and A- B.
     
  12. May 18, 2014 #11
    Thanks.

    One more question. Is the ##i## term meant to be present here:

    ##x_c = e^{t}(Ccos2t + Disin2t)## ?
     
  13. May 18, 2014 #12

    LCKurtz

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    Include the ##i## in the ##D##. While it appears the constants C and D may be complex, if the DE has real coefficients and real boundary conditions, they will come out real. If you left your answer in the ##\{e^{2it},e^{-2it}\}## form the constants would come out complex.
     
  14. May 18, 2014 #13
    Oh I see, thanks a lot!
     
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