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  1. F

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

    Number 2 is right. Number 1 isn't. y'' - 10y' + 25y = e^5t y(0)=0 y'(0)=0 L \{y^{\prime \prime}\} -10L \{y^\prime\} + 25L \{y\} = L \{e^{5t}\} s^2Y(s) -sy^\prime(0) - y(0) -10(sY(s) -y(0)) +25Y(s)= \frac{1}{s-5} s^2Y(s) -10sY(s) +25Y(s)= \frac{1}{s-5} Y(s)(s^2 -10s +25)=...
  2. F

    Laplace little differet one

    Also, L \{ y^{(n)} \} where n is the derivative is s^nY(s)-s^{n-1}y^{(n-1)}(0)-s^{n-2}y^{(n-2)}(0)-...-y(0) PS. click on the equation to see how to use LaTex extensions. A little popup box should appear with a "click to read LaTex guide" at the bottom. Good luck.
  3. F

    Laplace little differet one

    As with your other question, Laplace transforms are linear thus you can do each elemnt individually. What have you done thus far?
  4. F

    Inverse Laplace Transformation Problem

    pull out the e^{-s} leaving: L^{-1}{ \{ \frac{s}{s^2+1} \}=f(t-a) Now, the e can be converted to a unit step function U(t-a), and f(s-a) should be apparent. Combine the unit step function with F(s) to get an end result.
  5. F

    Laplace transform: where am I messing up?

    Got it. Thanks. How silly of me.
  6. F

    Laplace transform: where am I messing up?

    Ok, using the definition of Laplace transforms to find \L\{f(t)\} Given: f(t)=\{^{\sin{t}, 0\le{t}<{\pi}}_{0, t\ge{\pi}} So, this is what I did: \L\{\sin t\}=\int^{\pi}_{0} e^{-st}\sin t dt+\int^{\infty}_{\pi} e^{-st}(0)dt =\int^{\pi}_{0} e^{-st}\sin t dt...