Laplace Transform & Initial Value Problem

Question

$$ y'' + 4y = \begin{cases} t, & 0 \leq t < 3\\ 1, & 3 \leq t <\infty \end{cases} $$ $$y(0)=0, y'(0)=0$$

I need to find the Laplace transform of the solution of the given IVP above.

I don't know how to take Laplace transform of that kind of piecewise continuous function, need some help. Thanks.

Answer 1

Definition of Laplace transform is: $$F(s) = \int_0^\infty f(s) e^{-st}\, dt$$

Substitute the function and because integral is linear operation, it can be split into two integrals:

$ F(s) = \int_0^\infty \left(\begin{cases} t, & 0 \leq t < 3\\ 1, & 3 \leq t <\infty \end{cases} \right)e^{-st}\, dt = \int_0^3 t e^{-st}\, dt+\int_3^\infty e^{-st}\, dt $

Answer 2

You can define a piecewise function using the Heaviside Unit Step Function and then take the Laplace Transform of that.

In your example, we have:

$$\tag 1 f(t) = t - (t - 1)u(t-3)$$

The Laplace Transform of $(1)$ is given by:

$$\mathscr{L} (t - (t - 1)u(t-3)) = \dfrac{1}{s^2} + \dfrac{e^{-3s}}{s} -\dfrac{e^{-3s}(3s +1)}{s^2}$$

• $\mathscr{L} (y''(t)) = s^2y(s) -s y(0) -y'(0) = s^2y(s)$
• $\mathscr{L} (4y'(t)) = 4(sy(s) -y(0)) = 4s y(s)$

Can you know finish it off?

Answer 3

Hint:

Let $ Y(s):= \int_0^\infty y(t) e^{-st}\, dt$,

and we have $t[u(t)-u(t-3)]+u(t-3)=\begin{cases} t, & 0 \leq t < 3\\ 1, & 3 \leq t <\infty \end{cases}$

Then, apply Laplace to both sides.

Answer 4

Another approach may be given by:

If $\mathfrak{U}_c(x)= \left\{ \begin{array}{ll} 0 & 0\leq x \leq c \\ x & \quad x > c \end{array} \right.~~$ where $c>0$ is the Heaviside step function then:

$$ f(x) = \left\{ \begin{array}{ll} f_1(x) & \quad 0\leq x<a \\ f_2(x) & \quad x \ge a \end{array} \right. $$ can be written as $f(x)=f_1(x)-f_1(x)\mathfrak{U}_a(x)+f_2(x)\mathfrak{U}_a(x)$.

Apply this useful hint for your function on the right.