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 2 years ago
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) \[F(p)=\frac{p^2}{(p+3)^3}\]
 2 years ago
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) \[F(p)=\frac{p^2}{(p+3)^3}\]

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

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

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

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

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

mahmit2012
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1353613382376:dw

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.0i still dont understand this
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