These notes are set so that you get to prove the main results by solving smaller problems that when put together give the big result. The answers to the problems are in the videos. You will get the most out of these notes if you do (or try) the problems before looking at the videos.

More in inverse Laplace transforms.



The definition on the Laplace transform is $$ F(s) = {\mathcal L}\{ f(t)\} = \int_0^\infty e^{-st} f(t)\,dt. $$ With this notation we have shown $$ {\mathcal L}\{ e^{at}f(t)\} = F(s-a). $$ In term of inverse Laplace transforms this reads as $$ {\mathcal L}^{-1}\{F(s\} = e^{at} f(t). $$ Here we give examples of using this just in the special case of the following transforms, but the method works in much more generality.

\(f(t)\) \(F(s)\)
\(t^n\) \(\dfrac{n!}{s^{n+1}}\)
\(\sin(\omega t)\) \( \dfrac{\omega}{s^2+\omega^2}\)
\(\cos(\omega t)\) \( \dfrac{s}{s^2+\omega^2}\)


As an example of using these rules let us find the inverse transform of $$ G(s) = \frac{7}{(s-13)^4} . $$ This can be rewritten as $$ G(s) = \frac{7}{3!} \frac{3!}{(s-13)^4} = \frac{7}{3!} F(s) $$ where $$ F(s) = {\mathcal L}\{ t^3\} $$ and therefore $$ g(t) = {\mathcal L}^{-1}\{ G(s)\} = \frac{7}{3!} e^{13t} t^3 = \frac{7t^3 e^{13t}}{6} . $$

Problem: Find the inverse Laplace of $$ F(s) = \frac{3s + 2}{(s+6)^4} $$


Solution.



Problem: Find the inverse Laplace transform of $$ F(s) = \frac{3 s + 7}{2s^2+10} . $$


Solution.

Let us finish this up by doing one that requires completing the square. Rather that using the adding and subtracting trick that is usually use to motivate the method, at some point it just becomes easier to remember the formula $$ x^2 + bx = \left( x+ \frac{b}{2} \right)^2 - \left(\frac{b}2\right)^2. $$

Problem: Find the inverse Laplace transform of $$ F(s) = \frac{2s+3}{5s^2+20s+30} $$


Solution.