# 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?

The Mistake

$$X_1(s) = \frac{1}{s+4} \cdot(U(s)-X_2(s)) \xrightarrow{\mathscr{L}^{\,-1}} x_1(t) = e^{-4t}\cdot(u(t)-x_2(t))$$

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)

\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}