Question

Find the Laplace transform of the function given by f(t)=2t if t<=2 and f(t)=4 if t>2

Answer 1

Ok. So you split up your integrand into two parts, right?

Answer 2

I got 2+(2/(1+s)) and it was wrong

Answer 3

@Herp_Derp yes I did

Answer 4

\[L\{f(t)\}=\int_{0}^{inf}e^{-st}f(t)~dt\]

Answer 5

in this case, you split the integral up into 2 sections

Answer 6

I did that and I was wrong :(

Answer 7

can you type up your work? then we can see where the mistake might have happened

Answer 8

theres a shifting method that i can never remember, so i tend to do it the splitted way

Answer 9

yea 2\[2\int\limits_{0}^{2}e^(-st)+\int\limits_{0}^{infinity}4e^(-st)dt\]

Answer 10

it's supposed to be e^(-st)

Answer 11

0,2; then 2,inf for starters; and dont forget your "t" on the left side

Answer 12

f(t)=-2e^(-2s)/s + 2/s

Answer 13

oh sorry i actually have that down on my paper but I just typed it wrong

Answer 14

\[2\int_{0}^{2}e^{-st}t+4\int_{2}^{inf}e^{-st}dt\]

Answer 15

\[\large \int_0^2 2te^{-st}\,dt+\int_0^{\infty}4e^{-st}\,dt=2\int_0^2 te^{-st}\,dt+4\int_0^{\infty}e^{-st}\,dt\]Just making it look better :)

Answer 16

@amistre64 thats what I have. What do I do afterwards but thats where I went wrong

Answer 17

Thanks @Herp_Derp but I have that but afterwards is where I'm confused

Answer 18

2 to inf on the last one, the long right term is simple enough; the left side is by parts

Answer 19

I know I got that but afterwards is where I'm confused

Answer 20

for the firsr part the integral ends up equaling (3(1-e^(-2s)(2s+1))/(s^2)

Answer 21

and the second integral is (e^(-2s))/s

Answer 22

so (3(1-e^(-2s)(2s+1))/(s^2) +(e^(-2s))/s

Answer 23

\[2\int_0^2 te^{-st}\,dt+4\int_2^{\infty}e^{-st}\,dt\]
\[2(~\left.\frac{te^{-st}}{-s}\right|_{t=0}^{t=2}+\frac{1}{s}\left(\int_0^2 e^{-st}\,dt\right)+4\left.\frac{e^{-st}}{-s}\right|_{t=2}^{t=inf}\]

Answer 24

this thing aint latex friendly :)

\[2\left.\frac{te^{-st}}{-s}\right|_{t=0}^{t=2}+\frac{2}{s}\left(\int_0^2 e^{-st}\,dt\right)+4\left.\frac{e^{-st}}{-s}\right|_{t=2}^{t=inf}\]

\[2\left.\frac{te^{-st}}{-s}\right|_{t=0}^{t=2}+\left.\frac{2}{s}\frac{e^{-st}}{-s}\right|_{t=0}^{t=2}+4\left.\frac{e^{-st}}{-s}\right|_{t=2}^{t=inf}\]

\[2\frac{te^{-2s}}{-s}-2\frac{te^{0}}{-s}+\frac{2}{s}\frac{e^{-2s}}{-s}-\frac{2}{s}\frac{e^{0}}{-s}+4\frac{e^{-s(inf)}}{-s}-4\frac{e^{-2s}}{-s}\]

Answer 25

do we stop there?

Answer 26

id clean it up a little if i was you

Answer 27

ok thanks!

Answer 28

\[-\frac{2te^{-2s}}{s}+\frac{2t}{s}-\frac{2e^{-2s}}{s^2}+\frac{2}{s^2}+\cancel{4\frac{e^{-s(inf)}}{-s}}+\frac{4e^{-2s}}{s}\]

stuff like that

Answer 29

it's wrong :(

Answer 30

great time for loading issues eh

Answer 31

\[2\int_{0}^{2}e^{-st}t+4\int_{2}^{inf}e^{-st}dt\]
\[2(\frac{e^{-2s}-1}{s^2})+4\frac{e^{-2s}}{s}\]
or some variation of that should work

Answer 32

i never did get into the heaviside stuff to be competent enough in that method

Answer 33

sorry openstuy froze up! an it's still wrong