Laplace transformations
D Deleted member 4993 Guest Oct 28, 2021 #2 robbiej02 said: View attachment 29429 Click to expand... Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem.
robbiej02 said: View attachment 29429 Click to expand... Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem.
H HallsofIvy Elite Member Joined Jan 27, 2012 Messages 7,763 Oct 31, 2021 #3 Do you know that the Laplace Transform of f(x) is defined as \(\displaystyle F(t)= \int_0^\infty f(s)e^{-st}ds\)? With this f that is \(\displaystyle \int_0^2 -t e^{-st}dt\). To integrate that, use "integration by parts" with u= -t and \(\displaystyle dv= e^{-st}\). du= -dt v= -(1/s)e^{-st} \(\displaystyle (t/s)e^{-st}|_0^2+ (1/s)\int_0^2 e^{-st}dy\) \(\displaystyle (2/s)e^{-2s}- (1/s^2)e^{-st}|_0^2\) \(\displaystyle (2/s)e^{-2s}- (1/s^2)e^{-2s}- 1/s^2\)
Do you know that the Laplace Transform of f(x) is defined as \(\displaystyle F(t)= \int_0^\infty f(s)e^{-st}ds\)? With this f that is \(\displaystyle \int_0^2 -t e^{-st}dt\). To integrate that, use "integration by parts" with u= -t and \(\displaystyle dv= e^{-st}\). du= -dt v= -(1/s)e^{-st} \(\displaystyle (t/s)e^{-st}|_0^2+ (1/s)\int_0^2 e^{-st}dy\) \(\displaystyle (2/s)e^{-2s}- (1/s^2)e^{-st}|_0^2\) \(\displaystyle (2/s)e^{-2s}- (1/s^2)e^{-2s}- 1/s^2\)
