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Find Laplace Transform of (e^t + 2t^4 - cos (4t) + 10)

  1. Apr 22, 2004 #1
    Find Laplace Transform of...

    (e^t + 2t^4 - cos (4t) + 10)

    Thank you
     
  2. jcsd
  3. Apr 22, 2004 #2
    Remember that the Laplace Transform is a linear operator. That means that you can treat each one of the four terms as a separate little problem and then add them all back up at the end.

    Other than that, just apply the definition of the Laplace Transform or look at a table of transforms.

    cookiemonster
     
  4. Apr 22, 2004 #3
    But I don't know how to apply what you said to this problem. Can you please show me how?
     
  5. Apr 22, 2004 #4

    Tom Mattson

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    He meant that when calculating:

    L{et+2t4-cos(4t)+10},

    you can use the fact that the Laplace operator is linear to get:

    L{et}+2L{t4}-L{cos(4t)}+L{10}.

    That is, you can use linearity to turn your big problem into 4 small ones.

    edit: fixed a superscript bracket
     
  6. Apr 22, 2004 #5
    Is this right

    e



    1/(s-1) + 2* 4!/s^5- s/(s^2+16) + 10/s

    am I done?

    Thanks
     
  7. Apr 22, 2004 #6

    Tom Mattson

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    I get the same thing.
     
  8. Apr 23, 2004 #7
    Hi I just did two more Laplace problems can anyone please tell me if they're correct.

    1.Solve the following intital value problem using the method of Laplace transforms.

    y'' - 10y' + 25y = e^5t y(0)=0 y'(0)=0

    y"- 10y'+ 25y= e5t becomes -y'(0)- sy(0)+ L(y)- 10f(0)+ 10L(y)+ 25L(y)= L(e5x).

    Since y'(0)= y(0)= 0, this is just 34L(y)= L(e5x) so that
    L(y)= L(e5x)/34 and, since L(cf)= cL(f) for any constant c,
    y(x)= e5x/34.

    2.Find Laplace transform inverse

    L^-1 (4/s + 1/(s-1)^2 + (3s-16)/(s^2+64)


    4 +te^t + 3cos8t + -2sin8t

    Thank you
     
  9. Apr 23, 2004 #8
    Number 2 is right.

    Number 1 isn't.

    y'' - 10y' + 25y = e^5t y(0)=0 y'(0)=0

    [tex]L \{y^{\prime \prime}\} -10L \{y^\prime\} + 25L \{y\} = L \{e^{5t}\}[/tex]

    [tex]s^2Y(s) -sy^\prime(0) - y(0) -10(sY(s) -y(0)) +25Y(s)= \frac{1}{s-5}[/tex]

    [tex]s^2Y(s) -10sY(s) +25Y(s)= \frac{1}{s-5}[/tex]

    [tex]Y(s)(s^2 -10s +25)= \frac{1}{s-5}[/tex]

    [tex]Y(s)(s-5)^2= \frac{1}{s-5}[/tex]

    [tex]Y(s)= \frac{1}{(s-5)^3}[/tex]

    [tex]L^{-1}\{Y(s)\}=L^{-1} \{\frac{1}{(s-5)^3}\}[/tex]

    [tex]y(t)=\frac{1}{2}t^2e^{5t}[/tex]

    The above should be correct.
    Well anyway, good Luck and I hope it helps.
     
  10. Apr 24, 2004 #9
    Hi,

    I have a couple of more that I did. Could someone please check and see if these are right.
    1. FIND THE LAPLACE Transform of the unknown solution function for the following initial value problem:

    y" + 4y' - 5y = te^(t), y(0) = 1, y'(0)=0

    (Do Not actually find the function, only its transform. Then, without carrying out the steps, inidicate briefly how you would proceed to find the solution function.)

    s^2L = s + 4sL - 4 - 5L = Lte^t = 1/(s-1)^2

    (s^2 + 4s - 5)l = 1/(s-1)^2 + s +4 = (s^3+ 2s^2-7s+4)/(s-1)^2

    simplify
    = (s-1)^2(s+4)/(s-1)^3(s+5) = s=4/(s-1)(s+5)

    at this point I would do partial fractions.

    2.Find Laplace Inverse

    L^-1 { (s+10)/(s^2+8s+20)}

    s+10)/(s^2+8s+20) = (s+10)/(s^2+8s+16+4) = (s+10)/((s+4)^2+4)

    (s+10)/((s+4)^2+4) = (s-(-4))+ 3*2)/(s-(-4))^2+2^2) = (s-(-4)/((s-(-4))^2+2^2) + 3*2/(s-(-4)+2^2)

    The Inverse Laplace Transform of that is

    e^-4t cos 2t + 3 e^-4t sin 2t.

    3. Find Laplance Transform
    Find L (f(t)} where f(t) = e^-t (0 <= t < 5)
    -1 (t >= 5)

    For t in [0,5) the solution would be Int((e^{-t})(e^{-st},0,infinity,dt)=Int(e^{-t(s+1)},0,infinity,dt}=-e^{-t(s+1)}/(s+1)|0,infinity) = 1/(s+1).

    For t in [5,infinity) Int(-1*e^{-st},0,infinity,dt} =(1/s)e^{-st}|0,infinity)=-1/s.

    In piecewise notation you write L(f(t))={1/s, 0 <= t < 5 and -1/s, t >= 5}


    4. Find Laplace Inverse of

    L^-1 { (3s-4)/s(s-4) }

    so

    3s-4)/s(s-4) = A/s + B/(s-4)

    s*(3s-4)/(s(s-4)) = A + B*s/(s-4)

    Taking limit ( s to 0)

    -4/-4 = A

    A =1

    And

    (s-4)*(3s-4)/(s(s-4)) = A*(s-4)/s + B

    Taking limit ( s to 4)

    8/4 = B = 2

    so

    (3s-4)/s(s-4) = 1/s + 2/(s-4)





    Thank you
     
    Last edited: Apr 24, 2004
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