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ranyai12
Group Title
Find the Laplace transform of the function given by f(t)=2t if t<=2 and f(t)=4 if t>2
 one year ago
 one year ago
ranyai12 Group Title
Find the Laplace transform of the function given by f(t)=2t if t<=2 and f(t)=4 if t>2
 one year ago
 one year ago

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Herp_Derp Group TitleBest ResponseYou've already chosen the best response.0
Ok. So you split up your integrand into two parts, right?
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
I got 2+(2/(1+s)) and it was wrong
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
@Herp_Derp yes I did
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
\[L\{f(t)\}=\int_{0}^{inf}e^{st}f(t)~dt\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
in this case, you split the integral up into 2 sections
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
I did that and I was wrong :(
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
can you type up your work? then we can see where the mistake might have happened
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
theres a shifting method that i can never remember, so i tend to do it the splitted way
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
yea 2\[2\int\limits_{0}^{2}e^(st)+\int\limits_{0}^{infinity}4e^(st)dt\]
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
it's supposed to be e^(st)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
0,2; then 2,inf for starters; and dont forget your "t" on the left side
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
f(t)=2e^(2s)/s + 2/s
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
oh sorry i actually have that down on my paper but I just typed it wrong
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
\[2\int_{0}^{2}e^{st}t+4\int_{2}^{inf}e^{st}dt\]
 one year ago

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.0
\[\large \int_0^2 2te^{st}\,dt+\int_0^{\infty}4e^{st}\,dt=2\int_0^2 te^{st}\,dt+4\int_0^{\infty}e^{st}\,dt\]Just making it look better :)
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
@amistre64 thats what I have. What do I do afterwards but thats where I went wrong
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
Thanks @Herp_Derp but I have that but afterwards is where I'm confused
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
2 to inf on the last one, the long right term is simple enough; the left side is by parts
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
I know I got that but afterwards is where I'm confused
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
for the firsr part the integral ends up equaling (3(1e^(2s)(2s+1))/(s^2)
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
and the second integral is (e^(2s))/s
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
so (3(1e^(2s)(2s+1))/(s^2) +(e^(2s))/s
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
\[2\int_0^2 te^{st}\,dt+4\int_2^{\infty}e^{st}\,dt\] \[2(~\left.\frac{te^{st}}{s}\right_{t=0}^{t=2}+\frac{1}{s}\left(\int_0^2 e^{st}\,dt\right)+4\left.\frac{e^{st}}{s}\right_{t=2}^{t=inf}\]
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
this thing aint latex friendly :) \[2\left.\frac{te^{st}}{s}\right_{t=0}^{t=2}+\frac{2}{s}\left(\int_0^2 e^{st}\,dt\right)+4\left.\frac{e^{st}}{s}\right_{t=2}^{t=inf}\] \[2\left.\frac{te^{st}}{s}\right_{t=0}^{t=2}+\left.\frac{2}{s}\frac{e^{st}}{s}\right_{t=0}^{t=2}+4\left.\frac{e^{st}}{s}\right_{t=2}^{t=inf}\] \[2\frac{te^{2s}}{s}2\frac{te^{0}}{s}+\frac{2}{s}\frac{e^{2s}}{s}\frac{2}{s}\frac{e^{0}}{s}+4\frac{e^{s(inf)}}{s}4\frac{e^{2s}}{s}\]
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
do we stop there?
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
id clean it up a little if i was you
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
ok thanks!
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{2te^{2s}}{s}+\frac{2t}{s}\frac{2e^{2s}}{s^2}+\frac{2}{s^2}+\cancel{4\frac{e^{s(inf)}}{s}}+\frac{4e^{2s}}{s}\] stuff like that
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
it's wrong :(
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
great time for loading issues eh
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
\[2\int_{0}^{2}e^{st}t+4\int_{2}^{inf}e^{st}dt\] \[2(\frac{e^{2s}1}{s^2})+4\frac{e^{2s}}{s}\] or some variation of that should work
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
i never did get into the heaviside stuff to be competent enough in that method
 one year ago

ranyai12 Group TitleBest ResponseYou've already chosen the best response.0
sorry openstuy froze up! an it's still wrong
 one year ago
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