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Using laplace transforms to solve IVP problems with cramers rule

  1. Mar 12, 2012 #1
    1. The problem statement, all variables and given/known data
    y'' + 4y = sin(2t)

    y(0) = 1
    y'(0) = 1



    2. Relevant equations
    cramers rule


    3. The attempt at a solution

    after isolating Y(s) i end up with:

    Y(s) = 2 / (s^2+4)^2 + s/(s^2+4)

    after i do partial fraction decomposition, i get 4 equations with 4 unknowns, A, B, C and D

    i inserted B = 2- 4D into other 3 equations..

    when i used cramers rule to get values for A , B and C I am getting zeroes for all three variables. (??) ..also, I noticed that each equation also equals zero, but im not seeing what that might mean?
    i am doing something wrong

    my 3 equations are:

    A + 5C + D = 0
    -A -5C + D = 0
    2A + 16C -4D = 0

    any help is really appreciated...thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Mar 12, 2012
  2. jcsd
  3. Mar 13, 2012 #2

    vela

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    I don't get the same thing as you did for Y(s). Are both y(0) and y'(0) supposed to equal 1, or is one of them equal to 0?

    It would help if you showed what terms you're trying to break Y(s) into.
     
  4. Mar 13, 2012 #3
    Thanks for the reply. yup you're right i copied down one of the IV's wrong; should be y(0) = 1 and y'(0) = 0. But I actually solved with the correct IV's, so my Y(s) is the same:
    heres how i got it

    s^2Y(s) -sf(0) - f'(0) + 4Y(s) = 2/(s^+4)

    Y(s) = 2 / (s^2+4)^2 + s/(s^2+4)


    Y(s) = 2/(s^2+4)^2 + 2/(s^2+4)

    i decomposed just the first term on RHS into this

    2/(s^2+4)^2 = As+B/(s^2+4) + Cs+D/(s^2+4)^2

    multiplying by the denom i get:

    2 = As+B + Cs+D/(s^2+4)

    i next plugged in values of s=-1, 0, 1 and 2

    this gave me 4 equations and 4 unknowns

    i then solved for B in terms of D which is B=2-4D

    i am now left with 3 equations 3 unknowns

    A + 5C + D = 0
    -A -5C + D = 0
    2A + 16C + 4D = 0


    but like i said in initial post, i am puzzled as to what all the zeroes mean? (equations wrong somehow?)

    Also, a quick aside question: i just learned cramers rule for 3 equations and 3 unknowns does the rule work exactly the same for 4 equations with 4 unknowns? (and beyond?) and also, the method seems really susceptible to basic book keeping mistakes.. is this just because i just learned it, or is it not a very fast and efficient method for solving equations? thanks again

    UPDATE: i just plugged in my (originally) 4 equations into cramers rule calculator and got A=C=D=0 and B=2

    that would mean my decomposition = 2/s^2+4 + 2/s^2+4

    = sin2t + sin2t = 2sin2t
    but the book solution is cos2t - (1/4)tcos2t + (1/8) sin2t
     
    Last edited: Mar 13, 2012
  5. Mar 13, 2012 #4

    vela

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    You really need to get in the habit of using parentheses. You might notice that the LHS is exactly what you'd get when A=B=C=0 and D=2. In other words, it's already in the form you want.

    You made an algebra mistake. Since you multiplied by (s2+4)2, you should have
    $$2 = (As+B)(s^2+4) + Cs+D$$ If you solve for the coefficients, you'll indeed find that A=B=C=0 and D=2.
    Yes, Cramer's rule works for any number of unknowns. I find it's actually a nice method to use when you have unpleasant coefficients. For larger systems, it gets unwieldy simply because calculating determinants of large matrices is a pain. The largest system I'd use it for is probably one with three unknowns, unless I happened to have a lot of zero coefficients to make the big determinants easy to calculate.

    You will find it helpful to consider the derivative
    $$\frac{d}{ds} \left( \frac{2}{s^2+4} \right)$$ and use the property of the Laplace transform involving F'(s).
     
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