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## anonymous 3 years ago Find the Laplace transform of the function given by f(t)=2t if t<=2 and f(t)=4 if t>2

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1. anonymous

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

2. anonymous

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

3. anonymous

@Herp_Derp yes I did

4. amistre64

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

5. amistre64

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

6. anonymous

I did that and I was wrong :(

7. amistre64

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

8. amistre64

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

9. anonymous

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

10. anonymous

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

11. amistre64

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

12. anonymous

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

13. anonymous

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

14. amistre64

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

15. anonymous

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

16. anonymous

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

17. anonymous

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

18. amistre64

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

19. anonymous

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

20. anonymous

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

21. anonymous

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

22. anonymous

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

23. amistre64

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

24. amistre64

this thing aint latex friendly :) $2\left.\frac{te^{-st}}{-s}\right|_{t=0}^{t=2}+\frac{2}{s}\left(\int_0^2 e^{-st}\,dt\right)+4\left.\frac{e^{-st}}{-s}\right|_{t=2}^{t=inf}$ $2\left.\frac{te^{-st}}{-s}\right|_{t=0}^{t=2}+\left.\frac{2}{s}\frac{e^{-st}}{-s}\right|_{t=0}^{t=2}+4\left.\frac{e^{-st}}{-s}\right|_{t=2}^{t=inf}$ $2\frac{te^{-2s}}{-s}-2\frac{te^{0}}{-s}+\frac{2}{s}\frac{e^{-2s}}{-s}-\frac{2}{s}\frac{e^{0}}{-s}+4\frac{e^{-s(inf)}}{-s}-4\frac{e^{-2s}}{-s}$

25. anonymous

do we stop there?

26. amistre64

id clean it up a little if i was you

27. anonymous

ok thanks!

28. amistre64

$-\frac{2te^{-2s}}{s}+\frac{2t}{s}-\frac{2e^{-2s}}{s^2}+\frac{2}{s^2}+\cancel{4\frac{e^{-s(inf)}}{-s}}+\frac{4e^{-2s}}{s}$ stuff like that

29. anonymous

it's wrong :(

30. amistre64

great time for loading issues eh

31. amistre64

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

32. amistre64

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

33. anonymous

sorry openstuy froze up! an it's still wrong

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