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Laplace transformation to solve intial value problem

  1. Nov 28, 2008 #1
    1. The problem statement, all variables and given/known data
    laplace transformation to solve intial value problem
    y"-2y+y=-4e^1-t + 2t
    y(1)=0, y'(1)=-3


    2. Relevant equations



    3. The attempt at a solution
    K. I have one more on the review that I am unaware of where I need to start. So if anyone can point me in the right direction please feel free. My graduation depends on the final and I am so lost on how to start amny of these problems.... Thanks in advance
     
  2. jcsd
  3. Nov 28, 2008 #2

    rock.freak667

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    Take the Laplace transformation on both sides noting that L{y(t)}=Y(s), then just make Y(s) the subject, then find the inverse laplace transform of Y(s) and you will the solution.
     
  4. Nov 30, 2008 #3
    okay so I found the laplace of both sides and get

    2/s^3 + 2/s^2 + 1/s = (-4e(-t+1)/s) + 2t/s

    I'm not really sure where to go form here. I think that I want to set
    Y=L{y(t)}(s)
    L{y'}=sY
    L{y"}=s^2Y

    but I am not so sure that is the direction I need to head and I really am not sure were all this fits into the problem. Where do I head from here????
     
  5. Nov 30, 2008 #4

    rock.freak667

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    L{y(t)}=sY(s)-Y'(0)

    L{y''(t)}=s2Y(s)-sY(0)-Y'(0)

    L{ekt}= 1/(s-k)

    L{tn}=n!/sn+1
     
  6. Nov 30, 2008 #5
    ok I get the first two but why the second two in the above reply. I guess I don't know what I need to do with them...
     
  7. Nov 30, 2008 #6

    rock.freak667

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    On looking back to the question, you need to change the variable such that the conditions occur when t=0. To do this we let t=w+1

    so now we have

    y''(w-1)-2y'(w-1)+y(w+1)=4e-1-w-1+2(w+1)

    Now define u(w)=y(w+1)
    so now u'(w)=y'(w+1) and u''(w)=y''(w+1)

    Now when w=0 => u(0)=y(0+1)=y(1)=0 AND u'(w)=y'(1)=-3

    so now you need so solve the new equation

    u''-2u'+u=4e-w+2w+2


    Now take the Laplace transform on both sides of the equation.

    to elaborate on the second two...


    if you need to get L{e5t} it is simplt 1/(s-5)

    and L{t2}=2!/s3
     
  8. Dec 1, 2008 #7
    Okay so now I get
    L{u"}(s) = s2U(s) + 3
    L{u'}(s) = sU(s)


    u''-2u'+u=4e-w+2w+2

    taking the LT of both sides I get
    s2U(s)+3-2sU(s)+U(s)=(4/s+1)+(2/s2)+(2/s)
    (s2-2s+1)U(s)=(4/s+1)+(2/s2)+(2/s)-3

    I feel like I have messed up somewhere cause if I solve for U(s) then it gets really ugly....
     
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