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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)=...

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.

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

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.

Got it. Thanks. How silly of me.

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...