Convolution
Functions that are multiplied have a special relationship when the Laplace transform is taken.
The Essentials
The Laplace transform of two functions added together is the Laplace transform of the first added to the Laplace transform of the second. This isn't true for multiplication. Convolution is defined as:
\[
f(t)*g(t)=\int_0^tf(x)g(t-x)dx
\]
From the definition, this can be derived:
\[
\mathscr{L}(f(t)*g(t))=F(s)G(s)
\]
Example
Solve this equation. Express the answer using convolution:
\[
\begin{cases}
y''+y'-6y=f(t)\\
y(0)=0\\
y'(0)=0
\end{cases}
\]
First we take the Laplace transform:
\[
s^2Y(s)-sy(0)-y'(0)+sY(s)-y(0)-6Y(s)=F(s)
\]
\[
s^2Y(s)+sY(s)-6Y(s)=F(s)
\]
\[
Y(s)(s^2+s-6)=F(s)
\]
\[
Y(s)=\frac{F(s)}{(s^2+s-6)}
\]
Then, we can apply partial fraction expansion:
\[
Y(s)=F(s)\left(\frac{1/5}{s-2}-\frac{1/5}{s+3}\right)
\]
Now we can apply the convolution theorem:
\[
y(t) = f(t) * \left( \frac{1}{5} e^{2t} - \frac{1}{5} e^{-3t} \right)
\]