1. Motivation

Our main goal to learn the Laplace Transform is to apply it in initial-value problems:

\left\{\begin{array}{l} ay''+by'+cy=f(t), \\ y(0)=y_{0} \\ y'(0)=y_{1} \end{array}\right.

From what we learned before, it is clear that we can evaluate the Laplace Transform of f(t) and y(t) above:



Now, we want to be able to evaluate


such that we have an expression involving F(s).

2. Laplace transform derivatives theorem

3. Math behind Laplace transform of derivatives

a. First derivative

Let f(t) be a function in the variable t\geq0. Suppose f' is continuous for t\in[0,\infty). By definition:

\begin{aligned}\mathscr{L}\{f'(t)\} & = \int_{0}^{\infty}e^{-st}f'(t)dt \\ & = \lim_{b\to\infty}\int_{0}^{b}e^{-st}f'(t)dt \end{aligned}

Let u=e^{-st} and dv=f'(t)dt. Then




The integration by parts gives

\begin{aligned} \int_{0}^{b}e^{-st}f'(t)dt & = \left.e^{-st}f(t)\right|_{t=0}^{t=b}-\int_{0}^{b}-se^{-st}f(t)dt \\ & = e^{-sb}f(b)-e^{-s\cdot0}f(0)+s\int_{0}^{b}e^{-st}f(t)dt \\ & = e^{-sb}f(b)-f(0)+s\int_{0}^{b}e^{-st}f(t)dt \end{aligned}

We assume



\begin{aligned} \mathscr{L}\{f'(t)\} & = \lim_{b\to\infty}\left[e^{-sb}f(b)-f(0)+s\int_{0}^{b}e^{-st}f(t)dt\right] \\ & = \lim_{b\to\infty}e^{-sb}f(b)-f(0)+s\lim_{b\to\infty}\int_{0}^{b}e^{-st}f(t)dt \\ & = 0-f(0)+s\lim_{b\to\infty}\int_{0}^{b}e^{-st}f(t)dt \end{aligned}


\displaystyle F(s)=\lim_{b\to\infty}\int_{0}^{b}e^{-st}f(t)dt

we concluded that


b. Second derivative

In the same way, we can show a formula to the Laplace Transform of \mathscr{L}\{f''(t)\}. Assuming f' and f'' are both continuous on the interval [0,\infty), by the definition:


If u=e^{-st} then du=-se^{-st}dt and from dv=f''(t)dt we have v=f'(t). Applying integration by parts:

\begin{aligned} \int_{0}^{b}e^{-st}f''(t)dt & = \left.e^{-st}f'(t)\right|_{t=0}^{t=b}-\int_{0}^{b}-se^{-st}f'(t)dt \\ & = e^{-sb}f'(b)-e^{-s\cdot0}f'(0)+s\int_{0}^{b}e^{-st}f'(t)dt \\ & = e^{-sb}f'(b)-f'(0)+s\int_{0}^{b}e^{-st}f'(t)dt \end{aligned}

Again, we assume that


as b\to\infty. Thus,

\begin{aligned} \mathscr{L}\{f''(t)\} & = 0-f'(0)+s\cdot\overbrace{\lim_{b\to\infty}\int_{0}^{b}e^{-st}f'(t)d}^{\mathscr{L}\{f'(t)\}} \\ & = -f'(0)+s(sF(s)-f(0)) \end{aligned}



As we repeat the same idea showed before, we can get the theorem results.

4. Illustrative example

Applying the Laplace Transform to ODE in the initial-value problem, we get:


Due to the linearity of \mathscr{L} and since


we can rewrite as follows


Designing \mathscr{L}\{y(t)\}=Y(s) and using the formulas above, we have

\begin{aligned} \mathscr{L}\{y''(t)\} & = s^{2}Y(s)-sy(0)-y'(0) \\ & = s^{2}Y(s)-s\cdot1-(-1) \\ & = s^{2}Y(s)-s+1 \end{aligned}


\begin{aligned} \mathscr{L}\{y'(t)\} & = sY(s)-y(0) \\ & = sY(s)-1 \end{aligned}

Thus, the ODE becomes

\begin{aligned} & s^{2}Y(s)-s+1+3(sY(s)-1)=0 \\ & s^{2}Y(s)-s+1+3sY(s)-3=0 \\ & s^{2}Y(s)+3sY(s)-s-2=0 \end{aligned}

where the solution Y(s) is given by:

\displaystyle Y(s)=\frac{s+2}{s^2+3s}

Taking the inverse Laplace transform of Y(s), we can get the solution of the differential equation y(t).