I am just starting to learn convolutions and Laplace Transforms and am doing an example in my book and I don't understand why my methodology is incorrect so I will give the book's methodology first and then my methodology and I would appreciate if someone can tell me where I am misunderstanding something.
Find the PS to the following IDE:
$x'(t)+2x(t)-4\int_0^te^{(t-\tau)}x(\tau)d\tau = e^t$
where $x(0) = 0$
Book's Methodology:
let X(s) = $\scr{L}(x(t))$
Apply LaPlace Transform to both sides of the equation to get
$(sX(s)-x(0))+2X(s)-4\scr{L}(e^t\circledast x(t))=\scr{L}(e^t)$
We Know $ \scr{L}(e^t\circledast x(t)) = \scr{L}(e^t)\scr{L}(x(t))$
So let's also apply the IC's and rewrite the equation as
$sX(s)+2X(s)-4\cfrac{1}{s-1}X(s)=\scr{L}(e^t)$
Solving for X(s) and performing partial fraction, we obtain
$X(s) = \cfrac{1}{5}(\cfrac{1}{s-2}+\cfrac{4}{s+3})\scr{L}(e^t) $
Then applying inverse LT, we get
$x(t) = \cfrac{1}{5}\scr{L}^{-1}((\cfrac{1}{s-2}+\cfrac{4}{s+3})\scr{L}(e^t)) = \cfrac{1}{5}(e^{2t}+4e^{-3t}) \circledast e^t$
Using $e^{\alpha t} \circledast e^{\beta t} =\cfrac{e^{\alpha t}-e^{\beta t}}{\alpha - \beta}$
we get
$x(t) = \cfrac{1}{5}(\cfrac{e^{2t}-e^{t}}{2 - 1} + 4\cfrac{e^{-3t}-e^{t}}{-3 - 1}) = \cfrac{1}{20}(-3e^t+4e^{2t}-e^{-3t})$
Now What is wrong with my methodology?
My Methodology:
Apply LaPlace Transform to both sides of the equation to get
$(sX(s)-x(0))+2X(s)-4\scr{L}(e^t\circledast x(t))=\scr{L}(e^t)$
We Know $ \scr{L}(e^t\circledast x(t)) = \scr{L}(e^t)\scr{L}(x(t))$
So let's also apply the IC's and rewrite the equation as
$sX(s)+2X(s)-4\cfrac{1}{s-1}X(s)=\cfrac{1}{s-1}$ (Here I use $\cfrac {1}{s-1}$ instead on the RHS but I don't think there should be an issue here? Am I mistaken?)
$X(s)(\cfrac{s^2+s-6}{s-1}) = \cfrac{1}{s-1}$
$X(s) = \cfrac{(s-1)}{(s-1)(s^2+s-6)} = \cfrac{1}{(s^2+s-6)} = \cfrac{1}{5}(\cfrac{1}{s-2}) -\cfrac{1}{5}(\cfrac{1}{s+3})$ (I used partial fraction decomposition)
Now apply inverse laplace Transform
$x(t) = \cfrac{1}{5}\scr{L}^{-1}(\cfrac{1}{s-2})-\cfrac{1}{5}\scr{L}^{-1}(\cfrac{1}{s+3}) = \cfrac{1}{5}(e^{2t}-e^{-3t})$ which does not match the book's answer What am I doing wrong? Thank you for any help
They made a mistake in the book after this line: $$x(t) = \cfrac{1}{5}\left(\cfrac{e^{2t}-e^{t}}{2 - 1} + 4\cfrac{e^{-3t}-e^{t}}{-3 - 1} \right) $$ It should give: $$x(t) = \cfrac{1}{5}({e^{2t}\color{red}{-e^{t}}} -{e^{-3t} \color{red}{+e^{t}}})$$ This gives the same answer as yours: $$x(t) = \dfrac{1}{5}\left(e^{2t} -e^{-3t}\right)$$