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Second order differential

  1. Nov 28, 2008 #1
    1. The problem statement, all variables and given/known data
    I've completed this question but I dont think i've got it right. It's from a 2008 June FP1 further maths paper. There is a second part to the question I dont understand which I will post after I know I have got this part right :)

    Find, in terms of k, the general solution of the differential equation:

    [tex]\frac{d^{2}x}{dt} + 4 \frac{dx}{dt} + 3x = kt + 5[/tex]


    3. The attempt at a solution

    First of all i created auxilary equation

    [tex]T^{2} + 4T + 3[/tex]

    and solved it to get 2 real roots, 3 and 1

    2 real roots imply that the complementory soltuion:

    is [tex]y = Ae^{çx} + Be^{ßx}[/tex]
    i can fill in ç and ß, as it's the roots from the auxilary equation
    [tex]y = Ae^{x} + Be^{3x}[/tex]

    Now I do need use inspection to find the particular soltution:

    [tex]\frac{d^{2}x}{dt} + 4 \frac{dx}{dt} + 3x = kt + 5[/tex]


    let x = at + b

    [tex]\frac{d^{2}x}{dt} = 0 \frac{dx}{dt} = a[/tex]

    try:

    [tex] 0 + 4(a) + 3(at +b) \eqiv kt + 5[/tex]

    compare coefficients

    so 3a = k
    4a + 3b = 5

    a = k/3
    b = (15-4k)/3

    so the general solution is:

    [tex]y = \frac{k}{3} e^{x} + \frac{15-4k}{3} e^{3x}[/tex]

    is that right :) ?!

    Cheers :)
     
  2. jcsd
  3. Nov 28, 2008 #2

    tiny-tim

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    erm … -3 and -1 :redface:
     
  4. Nov 28, 2008 #3
    Sorry im being an idiot. I used comleting the square and by habbity i did

    (T-2)²
    when it should be +

    so is the final answer:

    [tex]
    y = \frac{k}{3} e^{-x} + \frac{15-4k}{3} e^{-3x}
    [/tex]
     
  5. Nov 28, 2008 #4

    tiny-tim

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    No! Try again.
    No, the particular solution is added to the general solution.
     
  6. Nov 28, 2008 #5
    so where does the complimentry solution come in?

    Thanks :)
     
  7. Nov 28, 2008 #6

    tiny-tim

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    oh … i meant complementary … i couldn't remember the word :redface:

    … the particular solution is added to the complementary solution.
     
  8. Nov 28, 2008 #7
    aha, cool.
    so
    [tex]
    y = Ae^{x} + Be^{3x}
    [/tex]
    +

    a = k/3
    b = (15-4k)/3

    is that it? How do I find the values of A and B

    Thanks :)
     
  9. Nov 28, 2008 #8

    tiny-tim

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    You don't need to :biggrin:

    the general solution does have two unknown constants.
     
  10. Nov 30, 2008 #9
    so is my final answer:

    [tex]

    y = Ae^{x} + Be^{3x} + \frac{k}{3} + \frac{15-4k}{3}

    [/tex]
     
  11. Nov 30, 2008 #10

    tiny-tim

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    No …

    see your post #3 …

    and i don't think the fractions are right either …

    please get into the habit of plugging your solution back into the original equation, to see if it works! :wink:

    (if it doesn't, then its wrong! :rolleyes:)
     
  12. Nov 30, 2008 #11

    Mark44

    Staff: Mentor

    No, your values for a and b, the coefficient of t and the constant, are wrong. They should be: a = k/3, b = 5/3 - 4k/9. And you forgot that the coefficients of x in the exponential terms are negative. (I have reverted to the variables of the original problem, x and t.)

    [Answer Removed]

    The homogeneous solution (aka complementary solution), [tex]x_h(t)[/tex] is the part of the general solution for which x''(t) + 4x'(t) + 3x(t) = 0. The particular solution is the part of the general solution for which x''(t) + 4x'(t) + 3x(t) = kt + 5.

    One thing about differential equations: getting the solutions takes a bit of doing, but if you know how to differentiate, you can check that your answer actually is a solution by verifying that the appropriate combination of x''(t), x'(t), and x(t) actually add up to what they're supposed to.
     
    Last edited by a moderator: Nov 30, 2008
  13. Nov 30, 2008 #12

    Mark44

    Staff: Mentor

    BTW, for future reference, differential equations are NOT precalculus!
     
  14. Nov 30, 2008 #13

    tiny-tim

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    Hey Mark …

    no complete solutions, please!

    could you edit? :smile:
     
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