Differential eqations course. Please see photo and show as much work as possible. 2) Use the definition of the Laplace Transform to show that L{te^-t} = 1/(s+1)^2 Solution Let y = te^(-t) L(s) = ?y*e^(-st)*dt Now, you need to combine the terms within the integral, then use integration by parts: t*e^(-t)*e^(-st) = t*e^(-t-st) L(s) = ?t*e^(-t-st)*dt from [0,Infinity] u = t dv = e^(-t-st)*dt du = dt v = (e^(-t-st))/(-1-s) L(s) = e^(-t-st)*t/(-1-s) - ?e^(-t-st)*dt/(-1-s) L(s) = e^(-t-st)*t/(-1-s) - e^(-t-st)/(-1-s)^2 [0,Infinity) L(s) = (0 - 0) - (0 - 1/(-1-s)^2)) L(s) = 1/(-1-s)^2 = 1/(1+s)^2.