Applying the Inverse Laplace Transform to both sides of an equation In the Laplace domain, we have this:
$$X_1(s) = \frac{1}{s+4}(U(s)-X_2(s))$$
I want to get this by:
$$\dot x_1(t) = -4x_1(t) + u(t)-x_2(t)$$
This is done by rearranging the first equation into:
$$(s+4)X_1(s) = U(s)-X_2(s)$$
Then take the Inverse Laplace Transform on both sides.
However, I would like know if I could achieve the same thing without rearranging the equation 1 first. Taking the Inverse Laplace Transform directly on both sides of the first equation resulting:
$$x_1(t) = e^{-4t}(u(t)-x_2(t))$$
Then take derivative with respect to $t$ on both sides:
$$\dot x_1(t) = -4e^{-4t}(u(t)-x_2(t)) + e^{-4t}(\dot u(t)-\dot x_2(t))$$
$$\dot x_1(t) = -4x_1 + e^{-4t}(\dot u(t)-\dot x_2(t))$$
I got stuck here. If this cannot be converted to the desired form, what is the reason for this?
 A: The Mistake
Your mistake is here:
$$
X_1(s) = \frac{1}{s+4} \cdot(U(s)-X_2(s)) \xrightarrow{\mathscr{L}^{\,-1}} x_1(t) = e^{-4t}\cdot(u(t)-x_2(t))
$$
In doing so, you have implicitly assumed the following equality as true:
$$
\color{red}{
\mathscr{L}^{-1} \left[ F(s) \cdot G(s) \right] = f(t) \cdot g(t)} \qquad \text{(Wrong!)}
$$

The Right Way
The correct version involves the convolution integral:
$$
\bbox[5px,border:1.1px solid black] {
\mathscr{L}^{-1} \left[ F(s) \cdot G(s) \right] = \left( {f * g} \right)\left( t \right) = \int_{{0}}^{{t}}{{f( {t - \tau } )g\left( \tau  \right)\,d\tau }}
}
$$
(Note that a convolution operation in time domain corresponds to a multiplication operation in Laplace complex frequency domain)
So, for your example:
$$
\begin{alignat}{1}
\mathscr{L}^{-1}[X_1(s)] = \mathscr{L}^{-1}\left[ \frac{1}{s+4} \cdot (U(s)-X_2(s)) \right]
\end{alignat}
$$
Then:
$$
\begin{alignat}{1}
x_1(t) = \int_{{0}}^{{t}}{{e^{-4(t - \tau)} \cdot \left( u(\tau) - x_2(\tau)  \right)\,d\tau }}
\implies
\dot{x}_1(t) &= \dfrac{d}{dt} \int_{{0}}^{{t}}{{e^{-4(t - \tau)} \cdot \left( u(\tau) - x_2(\tau)  \right)\,d\tau }}
\end{alignat}
$$
I don't think there is a way to get to the original differential equation from here without knowing nothing about $u(t)$ and $x_2(t)$. But if you choose two functions and solve this integral, it will lead to the same result given by the Inverse Laplace Transform operation without convolution.

An Example
Let's pick $u(t) = cos(t)$ and $x_2(t) = e^{-t}$. Then, we have $\mathscr{L} \left[u(t)-x_2(t) \right] = \dfrac{s-1}{s^3+s^2+s+1}$

*

*Anti-transforming without using convolution:

$$
X_1(s) = \dfrac{s-1}{(s+4)\cdot(s^3+s^2+s+1)} = \frac{-1/3\,}{s+1} + \frac{5/51}{s+4} + \frac{(4s+1)/17}{s^2+2}
\\[30px]
\begin{alignat}{1}
x_1(t) &= -\frac{1}{3}e^{-t} + \frac{5}{51}e^{-4t}+\frac{1}{17}\left(4\cos(t) + \sin(t) \right)
\\[5px]
\dot{x}_1(t) &= +\frac{1}{3}e^{-t} + \frac{20}{51}e^{-4t}+\frac{1}{17}\left(\cos(t) -4\sin(t) \right) 
\end{alignat}   
$$

*

*Anti-transforming using convolution:

$$
X_1(s) = \bigg(\frac{1}{s+4}\bigg) \cdot \bigg(\frac{s-1}{s^3+s^2+s+1}\bigg)
\\[30px]
\begin{alignat}{1}
\dot{x}_1(t) = \dfrac{d}{dt} \int_{{0}}^{{t}}{{e^{-4(t - \tau)} \cdot \left( \cos(\tau) - e^{-\tau}  \right)\,d\tau }} 
\,=\, \frac{1}{3}e^{-t} + \frac{20}{51}e^{-4t}+\frac{1}{17}\left(\cos(t) -4\sin(t) \right) 
\end{alignat}   
$$
