## 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:

F(s)=\mathscr{L}\{f(t)\}

Y(s)=\mathscr{L}\{y(t)\}

Now, we want to be able to evaluate

\mathscr{L}\{f''(t)\},\quad\mathscr{L}\{f'(t)\}

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

du=-se^{-st}dt

and

v=f(t)

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

\displaystyle\lim_{b\to\infty}e^{-sb}f(b)=0

then

\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}

Since

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

we concluded that

\mathscr{L}\{f'(t)\}=sF(s)-f(0)

### 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:

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

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

e^{-sb}f(b)\to0

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}

Therefore,

\mathscr{L}\{f''(t)\}=s^{2}F(s)-sf(0)-f'(0)

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:

\mathscr{L}\{y''(t)+3y'(t)\}=\mathscr{L}\{0\}

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

\mathscr{L}\{0\}=0

we can rewrite as follows

\mathscr{L}\{y''(t)\}+3\mathscr{L}\{y'(t)\}=0

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}

and

\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).