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If $$\mathscr{L}(y)=\frac{ne^{-pt_0}}{n^2+\omega^2}\left(\frac{1}{p+n}+\frac{n}{p^2+\omega^2}-\frac{p}{p^2+\omega^2}\right)$$ show that $$\bbox[yellow] {y=n\left(\frac{e^{-n(t-t_0)}}{n^2+\omega^2}+\frac{n\sin\Big(\omega (t - t_0)\Big)}{(n^2+\omega^2)\omega}-\frac{\cos \Big(\omega (t - t_0)\Big)}{n^2+\omega^2}\right)}$$ for $t\gt t_0$.


I'm having difficulty in reaching the yellow highlighted equation. I have included below the full question and answer for context:

In the above extract $Y=\mathscr{L}(y)$ and $L28, L2$ etc. are Laplace transforms from a table below where I have included only the relevant ones:

Laplace transform table


My attempt:

Multiplying out and taking the Inverse Laplace transform of each term $$y=\mathscr{L}^{-1}\left(\mathscr{L}(y)\right)=\mathscr{L}^{-1}\left(\frac{ne^{-pt_0}}{n^2+\omega^2}\left(\frac{1}{p+n}+\frac{n}{p^2+\omega^2}-\frac{p}{p^2+\omega^2}\right)\right)$$

$$=\mathscr{L}^{-1}\left(\color{red}{\frac{ne^{-pt_0}}{(p+n)}}\frac{1}{(n^2+\omega^2)}\right)+\mathscr{L}^{-1}\left(\frac{ne^{-pt_0}}{(p^2+\omega^2)}\frac{n}{(n^2+\omega^2)}\right)-\mathscr{L}^{-1}\left(\frac{ne^{-pt_0}}{(p^2+\omega^2)}\frac{p}{(n^2+\omega^2)}\right)$$

It was my understanding that the Inverse Laplace transform of $$\color{red}{\frac{ne^{-pt_0}}{p+n}}$$ is $$ne^{-n(t-t_0)}$$ by $L28$, but the problem is that this is only the numerator (without the $n$) of the first term in the bracket of the yellow highlighted equation:

$$\bbox[yellow] {y=n\left(\frac{e^{-n(t-t_0)}}{n^2+\omega^2}+\frac{n\sin\Big(\omega (t - t_0)\Big)}{(n^2+\omega^2)\omega}-\frac{\cos \Big(\omega (t - t_0)\Big)}{n^2+\omega^2}\right)}$$ So basically I don't understand how to find the Inverse Laplace transform of a product of $2$ functions. We are not using Convolution here otherwise the book extract would have mentioned this process; it's something much more simple than this that I am missing. Could someone please explain how to obtain the yellow highlighted equation?

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    $\begingroup$ Ignoring the constants, one of the factors is $e^{-pt_0}$. That spells translation. $\endgroup$ – Daniel Fischer Feb 20 '16 at 20:55
  • $\begingroup$ If $g(t) = f(t-h)$, how are the Laplace transforms of $f$ and $g$ related? $\endgroup$ – Daniel Fischer Feb 20 '16 at 20:59
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    $\begingroup$ No, the linearity means we have $\mathscr{L}(f_1 + f_2) = \mathscr{L}(f_1) + \mathscr{L}(f_2)$ and so on. $g$ is a translate of $f$, we define $g$ pointwise by setting $g(t) = f(t-h)$ for some $h\in \mathbb{R}$. (Besides, in your comment, where did $f$ go?) $\endgroup$ – Daniel Fischer Feb 20 '16 at 21:12
  • $\begingroup$ @Daniel Yes sorry; I meant to write $\mathscr{L}(f(t-h))=\mathscr{L}(f(t))-\mathscr{L}(f(h))$. I'm still confused about "if $g(t) = f(t-h)$, how are the Laplace transforms of $f$ and $g$ related?" $\endgroup$ – BLAZE Feb 20 '16 at 22:12
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$L28$ states that $$f(t) = \begin{cases} g(t-a) & \text{for} \quad t\gt a \gt 0 \quad\text{then}\quad \mathscr{L}(y)=e^{-pa}\times G(p)\\ 0 & \text{for} \quad t\lt a \end{cases}$$ Using the multiplied out version: $$\mathscr{L}^{-1}\left({\frac{ne^{-pt_0}}{(p+n)}}\frac{1}{(n^2+\omega^2)}\right)+\mathscr{L}^{-1}\left(\frac{ne^{-pt_0}}{(p^2+\omega^2)}\frac{n}{(n^2+\omega^2)}\right)-\mathscr{L}^{-1}\left(\frac{ne^{-pt_0}}{(p^2+\omega^2)}\frac{p}{(n^2+\omega^2)}\right)$$

Working through term by term, to get the inverse Laplace transform of the first of the three terms we rewrite $$\left({\frac{ne^{-pt_0}}{(p+n)}}\frac{1}{(n^2+\omega^2)}\right)$$ as $$\left(n\cdot{e^{-pt_0}}\cdot\color{#180}{{\frac{1}{(p+n)}}}\cdot\frac{1}{(n^2+\omega^2)}\right)$$ The objective is to match $${e^{-pt_0}}\cdot\color{#180}{{\frac{1}{(p+n)}}}\quad\text{with}\quad e^{-pa}\times G(p)$$ Where $a=t_0$, then $L2$ is used to get the inverse Laplace transform for the $\color{#180}{\mathrm{green}}$ part which becomes $e^{-nt}$. So we now have $${e^{-pt_0}}\cdot {e^{-nt}}\quad\text{to be matched with}\quad e^{-pa}\times G(p)$$ So if $$G(p) \quad\text{is}\quad e^{-nt}\quad\text{then by $L28$}\quad g(t-a)\quad\text{becomes}\quad e^{-n(t-t_0)}$$ with $a=t_0$

The $n$ and $\dfrac{1}{(n^2+\omega^2)}$ has no dependence on $p$ and are therefore just multiplying constants. Therefore the inverse Laplace transform of the first term by $L28$ is $$\mathscr{L}^{-1}\left(n\cdot {e^{-pt_0}}\cdot{{\frac{1}{(p+n)}}}\cdot\frac{1}{(n^2+\omega^2)}\right)={ne^{-n(t-t_0)}}\times\frac{1}{(n^2+\omega^2)}$$ as required.


In order to take the inverse Laplace transform of the second term $$\left(\frac{ne^{-pt_0}}{(p^2+\omega^2)}\frac{n} {(n^2+\omega^2)}\right)$$ we must first multiply top and bottom of $\dfrac{ne^{-pt_0}}{(p^2+\omega^2)}$ by $\omega$ to get $\dfrac{ne^{-pt_0}\omega}{(p^2+\omega^2)\omega}=\dfrac{n}{\omega}\cdot {e^{-pt_0}}\cdot \color{blue}{\dfrac{\omega}{p^2+\omega^2}}$ where the $\color{blue}{\mathrm{blue}}$ part is in the same form as $\dfrac{a}{p^2+a^2}$ as in $L3$ with $a=\omega$. So $\color{blue}{\dfrac{\omega}{p^2+\omega^2}}$ becomes $\sin(\omega t)$.

We now have using $L28\quad$ ${e^{-pt_0}}\cdot \sin(\omega t)\quad\text{to be matched with} \quad e^{-pa}\times G(p)$

So if $$G(p) \quad\text{is}\quad \sin(\omega t)\quad\text{then by $L28$}\quad g(t-a)\quad\text{becomes}\quad \sin\Big(\omega (t-t_0)\Big)$$ with $a=t_0$

The $\dfrac{n}{\omega}$ and again the $\dfrac{1}{(n^2+\omega^2)}$ have no dependence on $p$ and are therefore simply multiplying constants. Therefore the inverse Laplace transform of the second term by $L28$ is $$\mathscr{L}^{-1}\left(\frac{ne^{-pt_0}}{(p^2+\omega^2)}\frac{n}{(n^2+\omega^2)}\right)=\frac{n\sin\Big(\omega (t - t_0)\Big)}{(n^2+\omega^2)\omega}$$ as required.


For the last term we rewrite $$\left(\frac{ne^{-pt_0}}{(p^2+\omega^2)}\frac{p}{(n^2+\omega^2)}\right)\quad\text{as}\quad \frac{n}{(n^2+\omega^2)}\cdot e^{-pt_0}\cdot \color{red}{\frac{p}{(p^2+\omega^2)}}$$ where the $\color{red}{\mathrm{red}}$ part is in the same form as $\dfrac{p}{p^2+a^2}$ as in $L4$ with $a=\omega$. So $\color{red}{\dfrac{p}{(p^2+\omega^2)}}$ becomes $\cos(\omega t)$.

Similar to before; using $L28\quad$ ${e^{-pt_0}}\cdot \cos(\omega t)\quad\text{is to be matched with} \quad e^{-pa}\times G(p)$

So if $$G(p) \quad\text{is}\quad \cos(\omega t)\quad\text{then by $L28$}\quad g(t-a)\quad\text{becomes}\quad \cos\Big(\omega (t-t_0)\Big)$$ with $a=t_0$

As before; the $\dfrac{n}{(n^2+\omega^2)}$ has no dependence on $p$ and is therefore simply a multiplying constant. Therefore the inverse Laplace transform of the last term by $L28$ is $$\mathscr{L}^{-1}\left(\frac{ne^{-pt_0}}{(p^2+\omega^2)}\frac{p}{(n^2+\omega^2)}\right)=\frac{n\cos\Big(\omega (t - t_0)\Big)}{(n^2+\omega^2)}$$ as required.


Putting all the terms together and factoring out the $n$ we get $$y={ne^{-n(t-t_0)}}\times\frac{1}{(n^2+\omega^2)}+\frac{n\sin\Big(\omega (t - t_0)\Big)}{(n^2+\omega^2)\omega}-\frac{n\cos\Big(\omega (t - t_0)\Big)}{(n^2+\omega^2)}=\bbox[yellow] {n\left(\frac{e^{-n(t-t_0)}}{n^2+\omega^2}+\frac{n\sin\Big(\omega (t - t_0)\Big)}{(n^2+\omega^2)\omega}-\frac{\cos \Big(\omega (t - t_0)\Big)}{n^2+\omega^2}\right)}$$ as needed.

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