# How to Find the Laplace Transform of Derivatives and Integrals

concept

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.

fact

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

example

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. Now you may be wondering what to do in the case of higher order derivatives. Well just hold onto your pants because we're about to get there.

fact

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

example

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 \(\Laplace\{\frac{d^2}{dt^2}t^3\} = s^2 \frac{6}{s^4} - s\cdot 0 - 0 \) \( = \frac{6}{s^2} \)fact

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

example

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.
practice problems