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UnkleRhaukus
Group Title
Use partial fractions or the convolution theorem and the table of Laplace transforms, to find functions of \(t\) which have the following Laplace transforms.
(d) \[F(p)=\frac{p^2}{(p+3)^3}\]
 one year ago
 one year ago
UnkleRhaukus Group Title
Use partial fractions or the convolution theorem and the table of Laplace transforms, to find functions of \(t\) which have the following Laplace transforms. (d) \[F(p)=\frac{p^2}{(p+3)^3}\]
 one year ago
 one year ago

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UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
(d) partial fractions \[\begin{align*} F(p)&=\frac{p^2}{(p+3)^3} &=\frac{A}{p+3}+\frac B{(p+3)^2}+\frac C{(p+3)^3}\\ \\&&p^2=A(p+3)^2+B(p+3)+C\\ \\&\text{for } p=3&9=C\\ \\&\text{for } p=0&0=9A+3B+9\\ &&0=3A+B+3\\ &&B=3(A+1)\\ \\&\text{for } p=1&1=4A+2B+9\\ &&0=4A+2B+8\\ &&0=2A+B+4\\ &&0=2A3(A+1)+4\\ &&0=A+1\\ &&A=1\\ \\&&B=6\\ \\F(p) &=\frac{1}{p+3}\frac 6{(p+3)^2}+\frac 9{(p+3)^3}\\ \\ f(t)&=\mathcal L^{1}\left\{\frac{1}{p+3}\frac 6{(p+3)^2}+\frac 9{(p+3)^3}\right\}\\ &=e^{3t}\mathcal L^{1}\left\{\frac{1}{p}\frac 6{p^2}+\frac 9{p^3}\right\}\\ &=e^{3t}\left(16t+\frac 92t^2\right)\\ &=e^{3t}6te^{3t}+\frac 92t^2e^{3t}\\ \end{align*}\]
 one year ago

cnknd Group TitleBest ResponseYou've already chosen the best response.0
looks fine to me
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
(d) convolution theorem\[\begin{align*} F(p)&=\frac{p^2}{(p+3)^3}\\ &= p^2\times\frac12\times\frac{\Gamma(3)}{(p+3)^3}\\ \\ f(t)&=\tfrac12\mathcal L^{1}\left\{p^2\times\frac2{(p+3)^3}\right\}\\ &=\tfrac12\int\limits_0^t\delta''(t)\Big_{t\rightarrow tu}\left(u^2e^{3u}\right)\text du\\ &=\tfrac12\int\limits_0^t\delta''(tu)\left(u^2e^{3u}\right)\text du\\ &=\quad\dots\\ \end{align*}\]
 one year ago

cnknd Group TitleBest ResponseYou've already chosen the best response.0
actually, isn't there a known inverse transform for p/(p+3)^2?
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.0
i would like to add an alternate solution to partial fractions dw:1353607529536:dw\[\large \frac{ p^2 }{ (p+3)^3 }=\frac{ 1 }{ p+3 }\frac{ 6 }{ (p+3)^2 }+\frac{ 9 }{ (p+3)^3 }\]
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.0
also, in \[\frac{ p^2 }{ (p+3)^3 }=\frac{ A }{p+3 }+\frac{ B }{ (p+3)^2 }+\frac{ C }{ (p+3)^3 }\] LET \[\large \phi(p)=p^2\]\[\large C=\phi(3)=9\]\[\large B=\phi \prime (3)=6\]\[\large A = \frac{ \phi \prime \prime (3) }{ 2! }=1\]
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
dw:1353613382376:dw
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
i still dont understand this
 one year ago
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