Question

Please check my work!

Find the Laplace Transform of the given piecewise function:

Answer 1

\[\huge f(t)=\left\{\begin{matrix} 0, & t < 2\\ (t-2)^2, & t \ge 2 \end{matrix}\right.\]

\[\huge \mathcal{L}[(t-2)^2u_2(t)] =e^{-2s} \mathcal{L}[t^2]=2e^{-2s} \cdot \frac{1}{s^3}\]

Answer 2

@ganeshie8 @IrishBoy123 @freckles X)

Answer 3

Woops, the end got cut off :(

\[\huge 2e^{-2s} \cdot \frac{1}{s^3}\]

Answer 4

haven't looked at the table but that seems great

@Astrophysics

Answer 5

\[ f(t)=\left\{\begin{matrix} 0, & t < 2\\ (t-2)^2, & t \ge 2 \end{matrix}\right.\]

\[ \mathcal{L}[(t-2)^2u_2(t)] =e^{-2s} \mathcal{L}[t^2]=2e^{-2s} \cdot \frac{1}{s^3}\]

Answer 6

Okay, I was just checking! I make the dumbest mistakes I tell you.. hehe

Answer 7

Learning laplace as well eh :D, me too

Answer 8

@Astrophysics It's really not fun :( I didn't mind the other sections of differential eq's though!

Answer 9

Yeah! I feel exactly the same way!