Inverse Laplace Transform - Some Confusion on Breaking into Component Functions (Find the solution y(t) ) Been redoing the last part of the problem and not understanding the book solution.
Would appreciate clarification. Thanks.
Given: $\frac{d^2y}{dt^2}+\frac{2dy}{dt}+26y=\delta _4\left(t\right)\:$ $ -> IC.\:y\left(0\right)=1,\:y'\left(0\right)=0$
Did:

*

*apply$ :\ L[f(t)]\:$ to all $\ fcn(t)\:$

*apply derivative rules for Laplace Transforms

*get $:\ L[y(t)]\:$  in terms of $  \ fcn(s)\: $

*complex $\lambda \:$ --> complete the square in the denominator of $  \ fcn(s)\: $ to generate a known form $ : \sin\left(\omega t\right)\:or\:cos\left(\omega t\right)\ $
My basic $s $  domain expression  is Correct:
$ \ Y(s) =  \frac{e^{-4s}}{\left(s+1\right)^2+5^2}\:+\:\frac{\left(s+2\right)}{\left(s+1\right)^2+5^2}\ $
Book Solution: $\ y\left(t\right)=\:e^{-t}cos5t\:+\frac{1}{5}e^{-t}sin5t+\frac{1}{5}u_4\left(t\right)e^{-\left(t-4\right)}sin\left(5\left(t-4\right)\right) \ $
My "simplified" s domain expression
$ \ Y\left(s\right)=\frac{1}{5}\cdot e^{-4s}\cdot \frac{5}{\left(s+1\right)^2+5^2}+\frac{1}{5}\cdot \left(s+2\right)\cdot \frac{5}{\left(s+1\right)^2+5^2} \ $
My solution
$ \:y\left(t\right)=\:\delta _4\left(t\right)\cdot \frac{1}{5}e^{-t}\cdot sin\left(5t\right)+e^{2t}\cdot \frac{1}{5}\left(sin\left(5t\right)\right) \ $
Don't Understand:

*

*Where the heaviside function comes from.

*Why you can/have to split
the $ \ s+2\  $ numerator function using partial fractions.

 A: Recall that
$$\mathcal L^{-1}_t\left\{e^{-cs}F(s)\right\}=u_c(t)f(t-c)$$
$$\implies\mathcal L^{-1}_t\left\{\frac{e^{-4s}}{(s+1)^2+5^2}\right\}=u_4(t)f(t-4)$$
where $f(t)$ is the inverse Laplace transform of $F(s)$. Also, recall that
$$\mathcal L^{-1}_t\left\{F(s-c)\right\}=e^{ct}f(t)$$
$$\implies\mathcal L^{-1}_t\left\{\frac1{(s+1)^2+5^2}\right\}=e^{-t}\mathcal L^{-1}_t\left\{\frac1{s^2+5^2}\right\}=\frac15e^{-t}\sin(5t)$$
Putting these together gives
$$\mathcal L^{-1}_t\left\{\frac{e^{-4s}}{(s+1)^2+5^2}\right\}=\frac15u_4(t)e^{-(t-4)}\sin(5(t-4))$$

On the other hand,
$$\frac{s+2}{(s+1)^2+5^2}=\frac{s+1}{(s+1)^2+5^2}+\frac1{(s+1)^2+5^2}$$
and you should recognize these as having the inverse transforms (using one of the properties mentioned above),
$$\mathcal L^{-1}_t\left\{\frac{s+1}{(s+1)^2+5^2}\right\}=e^{-t}\mathcal L^{-1}_t\left\{\frac s{s^2+5^2}\right\}=e^{-t}\cos(5t)$$
$$\mathcal L^{-1}_t\left\{\frac1{(s+1)^2+5^2}\right\}=e^{-t}\mathcal L^{-1}_t\left\{\frac1{s^2+5^2}\right\}=\frac15e^{-t}\sin(5t)$$
