Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
UnkleRhaukus
Group Title
Is there a simpler method to show from the definition of the Laplace transform.
\[\mathcal L\left\{t\sin(nt)\right\}=\frac{2pn}{\left(p^2+n^2\right)^2}\]
 one year ago
 one year ago
UnkleRhaukus Group Title
Is there a simpler method to show from the definition of the Laplace transform. \[\mathcal L\left\{t\sin(nt)\right\}=\frac{2pn}{\left(p^2+n^2\right)^2}\]
 one year ago
 one year ago

This Question is Closed

LolWolf Group TitleBest ResponseYou've already chosen the best response.0
I haven't come across a nicer one, myself, but it'd be nice to see.
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.0
(If what I'm seeing is right, of course, considering it's cut off)
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.0
http://mathurl.com/aq279pu.png
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
so u can't use properties of LT ?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[\mathcal L\left\{t\sin(nt)\right\}=\int\limits_0^\infty t\sin(nt)e^{pt}\text dt\] \[=\int_0^\infty t\frac{e^{int}e^{int}}{2i}e^{pt}\text dt\] \[=\frac 1{2i}\int_0^\infty t({e^{int}e^{int}})e^{pt}\text dt\] \[=\frac 1{2i}\left(\int_0^\infty t\cdot e^{(pin)t}\text dt\int_0^\infty t\cdot e^{(p+in)t}\text dt\right)\] \[=\frac 1{2i}\left[\left(\left.\frac{t\cdot e^{(pin)}}{(pin)}\right_0^\infty\int_0^\infty\frac{e^{(pin)t}}{(pin)}\text dt\right)\right.\]\[\qquad\left.\left(\left.\frac{t\cdot e^{(p+in)}}{(p+in)}\right_0^\infty\int_0^\infty\frac{e^{(p+in)t}}{(p+in)}\text dt\right)\right]\] \[=\frac 1{2i}\left[\left(0+\int_0^\infty\frac{e^{(pin)t}}{(pin)}\text dt\right)\left(0+\int_0^\infty\frac{e^{(p+in)t}}{(p+in)}\text dt\right)\right]\] \[=\frac 1{2i}\left[\int_0^\infty\frac{e^{(pin)t}}{(pin)}\text dt\int_0^\infty\frac{e^{(p+in)t}}{(p+in)}\text dt\right]\] \[=\frac 1{2i}\left[\left.\frac{e^{(pin)t}}{(pin)^2}\frac{e^{(p+in)t}}{(p+in)^2}\right_0^\infty\right]\] \[=\frac 1{2i}\left[\frac{1}{(pin)^2}\frac{1}{(p+in)^2}\right]\] \[=\frac 1{2i}\left[\frac{(p+in)^2(p+in)^2}{(pin)^2(pin)^2}\right]\] \[=\frac 1{2i}\left[\frac{(p^22ipnn^2)(p^2+2ipnn^2)}{\left((pin)(p+in)\right)^2}\right]\] \[=\frac 1{2i}\left[\frac{4ipn}{\left(p^2+n^2\right)^2}\right]\] \[=\frac{2pn}{\left(p^2+n^2\right)^2}\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
the question was "From the definition, evaluate the following Laplace transforms."
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
0 L(sinnt)=n/p2+n2 L(tsinnt)=d/dp(n/p2+n2)=2np/(p2_n2)2
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.0
From definition...?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[\mathcal {L}\{ t f(t)\}=\frac{\text d}{\text dp}F(p)\] @mahmit2012 does that follow from the definition?
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.0
I guess you could write it as a Lemma... don't know how that will fly, with your prof. though.
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
Yes, that is correct.
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.0
I'm still thinking it requires some work, it doesn't just follow from definition.
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[\mathcal L\left\{t sin(nt) \right\}\] \[\qquad\qquad\text{let } f(t)=sin(nt)\]\[\qquad\qquad F(t)= \frac{n}{p^2+n^2}\]\[\qquad\qquad \mathcal {L}\{ t f(t)\}=\frac{\text d}{\text dp}F(p)\]\[\mathcal L\left\{tf(p)\right\}=\frac{\text d}{\text dp}\left(\frac{n}{p^2+n^2}\right)\]\[=\frac{\text d}{\text dp}\left({n}({p^2+n^2})^{1}\right)\]\[=2pn(p^2+n^2)^{2}\]\[=\frac{2pn}{(p^2+n^2)^2}\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
that is much quicker
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
but u didn't use definition.....
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.0
I agree, but I thought you said it had to be from definition...?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
u used a property that follows from definition
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
i did prove \[f(t)=sin(nt)\] \[F(p)= \frac{n}{p^2+n^2}\] in a earlier question
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
with the given question, i would go with your earlier response, though its longer
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.0
What I was trying to say, as with @hartnn is, just go ahead and prove this: \[\mathcal {L}\{ t f(t)\}=\frac{d}{dp}F(p)\] as a lemma, and you know that you're working from definition.
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
i havent proved \[\mathcal {L}\{ t f(t)\}=\frac{\text d}{\text dp}F(p)\] yet
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.0
Yes, and you should as it's not a corollary of the definition, it's actually a lemma. Once you've got that, just reference it, I haven't had professors nag me about doing such like this, yet, but you never know.
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.