Question

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

Answer 1

(d) partial fractions
\[\begin{align*}
F(p)&=\frac{p^2}{(p+3)^3}
&=\frac{A}{p+3}+\frac B{(p+3)^2}+\frac C{(p+3)^3}\\
\\&&p^2=A(p+3)^2+B(p+3)+C\\
\\&\text{for } p=-3&9=C\\
\\&\text{for } p=0&0=9A+3B+9\\
&&0=3A+B+3\\
&&B=-3(A+1)\\
\\&\text{for } p=-1&1=4A+2B+9\\
&&0=4A+2B+8\\
&&0=2A+B+4\\
&&0=2A-3(A+1)+4\\
&&0=-A+1\\
&&A=1\\
\\&&B=-6\\
\\F(p)
&=\frac{1}{p+3}-\frac 6{(p+3)^2}+\frac 9{(p+3)^3}\\
\\
f(t)&=\mathcal L^{-1}\left\{\frac{1}{p+3}-\frac 6{(p+3)^2}+\frac 9{(p+3)^3}\right\}\\
&=e^{-3t}\mathcal L^{-1}\left\{\frac{1}{p}-\frac 6{p^2}+\frac 9{p^3}\right\}\\
&=e^{-3t}\left(1-6t+\frac 92t^2\right)\\
&=e^{-3t}-6te^{-3t}+\frac 92t^2e^{-3t}\\
\end{align*}\]

Answer 2

looks fine to me

Answer 3

(d) convolution theorem\[\begin{align*}
F(p)&=\frac{p^2}{(p+3)^3}\\
&= p^2\times\frac12\times\frac{\Gamma(3)}{(p+3)^3}\\
\\
f(t)&=\tfrac12\mathcal L^{-1}\left\{p^2\times\frac2{(p+3)^3}\right\}\\
&=\tfrac12\int\limits_0^t\delta''(t)\Big|_{t\rightarrow t-u}\left(u^2e^{-3u}\right)\text du\\
&=\tfrac12\int\limits_0^t\delta''(t-u)\left(u^2e^{-3u}\right)\text du\\
&=\quad\dots\\
\end{align*}\]

Answer 4

actually, isn't there a known inverse transform for p/(p+3)^2?

Answer 5

i would like to add an alternate solution to partial fractions
\[\large \frac{ p^2 }{ (p+3)^3 }=\frac{ 1 }{ p+3 }-\frac{ 6 }{ (p+3)^2 }+\frac{ 9 }{ (p+3)^3 }\]