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.