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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
 2 years ago
 2 years 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
 2 years ago
 2 years 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?
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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