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UnkleRhaukus
 2 years ago
Is there a simpler method to show from the definition of the Laplace transform.
\[\mathcal L\left\{t\sin(nt)\right\}=\frac{2pn}{\left(p^2+n^2\right)^2}\]
UnkleRhaukus
 2 years ago
Is there a simpler method to show from the definition of the Laplace transform. \[\mathcal L\left\{t\sin(nt)\right\}=\frac{2pn}{\left(p^2+n^2\right)^2}\]

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

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

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.0so u can't use properties of LT ?

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

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0the question was "From the definition, evaluate the following Laplace transforms."

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.00 L(sinnt)=n/p2+n2 L(tsinnt)=d/dp(n/p2+n2)=2np/(p2_n2)2

UnkleRhaukus
 2 years ago
Best 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?

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

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, that is correct.

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0I'm still thinking it requires some work, it doesn't just follow from definition.

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

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0that is much quicker

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.0but u didn't use definition.....

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0I agree, but I thought you said it had to be from definition...?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.0u used a property that follows from definition

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

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.0with the given question, i would go with your earlier response, though its longer

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0What 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.

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

LolWolf
 2 years ago
Best ResponseYou've already chosen the best response.0Yes, 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.
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