How to Find the Laplace Transform of Derivatives and Integrals

If you want to take the Laplace transform of a differential equation it's going to be helpful to know how to transform a derivative. On the other hand, many circuit equations have integrals in them, so it'd be nice if we could turn that into a Laplace transform too. Well guess what? You're about to learn how to turn derivatives and integrals into algebraic equations. You lucky thing you.

For some \(f(t),\quad \Laplace\{\frac{df(t)}{dt}\} = sF(s) - f(0)\)

Find the Laplace transform of \(\frac{d}{dt}e^{10t}\)

We see that \(f(t) = e^{10t} \implies F(s) = \frac{1}{s-10}\) We also have \(f(0) = e^{10\cdot 0} = 1\) So \(\Laplace\{\frac{df(t)}{dt}\} = \frac{s}{s-10} - 1\) \( = \frac{10}{s-10}\) Which we can verify by doing the derivative and then taking the Laplace transform. \(\frac{d}{dt}e^{10t} = 10e^{10t}\) \(\Laplace\{10e^{10t}\} = \frac{10}{s-10}\) just as we calculated.

The formula for the Laplace transform of higher order derivatives is: \(\Laplace\{\frac{d^n}{dt^n}f(t)\} = s^n\Laplace\{f\} - s^{n-1}f(0) - s^{n-2}f'(0) - \cdots - f^{(n-1)}(0)\)

Find \(\Laplace\{\frac{d^2}{dt^2}t^3\}\)

\(f(t) = t^3,\quad f(0) = 0\) \(f'(t) = 3t^2,\quad f'(0) = 0\) \(\Laplace\{t^3\} = \frac{6}{s^2}\) So \(\Laplace\{\frac{d^2}{dt^2}t^3\} = s^2 \frac{6}{s^4} - s\cdot 0 - 0 \) \( = \frac{6}{s^2} \)

The laplace transform of an integral is given by: \( \Laplace\{\int_0^t f(t)dt\} = \frac{1}{s}F(s) \)

Find the inverse Laplace transform of \(\frac{1}{s(s-2)}\)

Rewriting it as \(\frac{1}{s}\frac{1}{s-2}\) we can see that this is \(\int_0^t e^{2t}dt\) Completing the integral we get: \(\int_0^t e^{2t}dt = \frac{1}{2}e^{2t}\evalat_0^t = \frac{1}{2}e^{2t} - \frac{1}{2}\) We can check by taking the Laplace transform of our answer and seeing that it matches up with the question. \(\Laplace\{\frac{1}{2}e^{2t} - \frac{1}{2}\} = \frac{1}{2(s-2)} - \frac{1}{2s}\) \( = \frac{2s - 2(s-2)}{4s(s-2)} = \frac{1}{s(s-2)} \) Just as we wanted.