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Differential Equations old and the new

  1. May 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Given:
    y''[t] + 25 y[t] = 0

    I know that the solution to this DE is of the form:

    y[t] = K1 E^(-5 i t) + K2 ​​E ^(5 i t)

    I get that, that makes sense to me, however when I look in old DE books I see the solution to the same problem written as:


    C1 Cos[5 t] + C2 Sin[k t]


    How do they get this, and how can I establish a transition between the two?
    How are they related?

    Thanks!
    -James
     
  2. jcsd
  3. May 24, 2013 #2

    Ray Vickson

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    Surely you must have seen ##e^{ix} = \cos x + i \sin x.## It is one of the most fundamental results you will ever meet.
     
  4. May 24, 2013 #3
    Ok, I guess it's my constants that are getting me in a twist.

    Where am I going wrong here?

    knowing that:
    E^(ix) = cos(x) + i sin(x)
    and
    E^(-ix) = cos(x) - i sin(x)

    K1 E^(-5 i t) + K2 ​​E ^(5 i t)
    (Let x = 5t)
    = K1 (cos(x) - i sin(x)) + K2(cos(x) + i sin(x))
    = cos(x) (K1 + K2) + i sin(x) (K2-k1)

    So what should I do from here?
    Or am I just really over thinking this one?
     
  5. May 24, 2013 #4

    Ray Vickson

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    So you have ##K_1+K_2 = C_1## and ##i (K_2 - K_1) = C_2.## You are over-thinking it: a linear combinations of sin(x) and cos(x) can be written as a linear combination of exp(ix) and exp(-ix). Either way of writing it is fine. And, BTW: there is nothing "new" about using complex exps and nothing "old" about using sin and cos: you just fit the form to the problem you are addressing. Some problems need the sin and cos form; others cry out to be left in exp form.
     
  6. May 24, 2013 #5
    oh ok.
    One final question regarding this.
    When I switch up the constants, I get
    C1 Cos(5t) + C2 Sin(5t)
    But, I'm suppose to come up with
    C1 Cos[5 t] + C2 Sin[k t]

    So what should I do about that k? Why is it not a 5? Or does k have to equal 5?

    Thanks.
    -James
     
  7. May 24, 2013 #6

    Ray Vickson

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    You tell me.
     
  8. May 24, 2013 #7
    Well, I want to say that:
    C1 Cos(5t) + C2 Sin(5t) = C1 Cos(5t) + C2 Sin(kt)
    and that k is just 5.
    Is that right to say?

    I'm asking this not because I have trouble determining if k = 5, but mainly to make sure I did my work correct in finding
    C1 Cos(5t) + C2 Sin(5t)
    then I know that k = 5.
    But then that just makes me confused because how would of they solved for it by having a k there and not knowing that k = 5.
     
  9. May 24, 2013 #8

    Ray Vickson

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    I don't get it: you know that sin(5t) and cos(5t) are solutions, so k must be 5. End of story.

    If you want to make a lot of unnecessary work for yourself, you can plug in sin(kt) into the DE and see if it works; it will work if you choose k properly.
     
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