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

Differential Equations
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(d) partial fractions \[\begin{align*} F(p)&=\frac{p^2}{(p+3)^3} &=\frac{A}{p+3}+\frac B{(p+3)^2}+\frac C{(p+3)^3}\\ \\&&p^2=A(p+3)^2+B(p+3)+C\\ \\&\text{for } p=-3&9=C\\ \\&\text{for } p=0&0=9A+3B+9\\ &&0=3A+B+3\\ &&B=-3(A+1)\\ \\&\text{for } p=-1&1=4A+2B+9\\ &&0=4A+2B+8\\ &&0=2A+B+4\\ &&0=2A-3(A+1)+4\\ &&0=-A+1\\ &&A=1\\ \\&&B=-6\\ \\F(p) &=\frac{1}{p+3}-\frac 6{(p+3)^2}+\frac 9{(p+3)^3}\\ \\ f(t)&=\mathcal L^{-1}\left\{\frac{1}{p+3}-\frac 6{(p+3)^2}+\frac 9{(p+3)^3}\right\}\\ &=e^{-3t}\mathcal L^{-1}\left\{\frac{1}{p}-\frac 6{p^2}+\frac 9{p^3}\right\}\\ &=e^{-3t}\left(1-6t+\frac 92t^2\right)\\ &=e^{-3t}-6te^{-3t}+\frac 92t^2e^{-3t}\\ \end{align*}\]
looks fine to me
(d) convolution theorem\[\begin{align*} F(p)&=\frac{p^2}{(p+3)^3}\\ &= p^2\times\frac12\times\frac{\Gamma(3)}{(p+3)^3}\\ \\ f(t)&=\tfrac12\mathcal L^{-1}\left\{p^2\times\frac2{(p+3)^3}\right\}\\ &=\tfrac12\int\limits_0^t\delta''(t)\Big|_{t\rightarrow t-u}\left(u^2e^{-3u}\right)\text du\\ &=\tfrac12\int\limits_0^t\delta''(t-u)\left(u^2e^{-3u}\right)\text du\\ &=\quad\dots\\ \end{align*}\]

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Other answers:

actually, isn't there a known inverse transform for p/(p+3)^2?
i would like to add an alternate solution to partial fractions |dw:1353607529536:dw|\[\large \frac{ p^2 }{ (p+3)^3 }=\frac{ 1 }{ p+3 }-\frac{ 6 }{ (p+3)^2 }+\frac{ 9 }{ (p+3)^3 }\]
also, in \[\frac{ p^2 }{ (p+3)^3 }=\frac{ A }{p+3 }+\frac{ B }{ (p+3)^2 }+\frac{ C }{ (p+3)^3 }\] LET \[\large \phi(p)=p^2\]\[\large C=\phi(-3)=9\]\[\large B=\phi \prime (-3)=-6\]\[\large A = \frac{ \phi \prime \prime (-3) }{ 2! }=1\]
|dw:1353613382376:dw|
i still dont understand this

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