Help with basic Laplace Transform - unsure of procedure!!!

Question

I am working on this Laplace Transform, and I've tried looking for a similar example off which to base my own work, but haven't been very successful. I'm confused by the formatting and don't know how to proceed:

Find the Laplace transform Y(s) = L{y} of the solution of the given initial value problem.

$y" + 4y = (1 [0,pi) ; 0 [pi, inf)$

$y(0) = 6$

$y'(0) = 7$

I've never encountered one like this (we just started this week), and I'm not entirely confident on how to proceed. Could someone walk me through a procedure or point me to a similar example or post?

Thanks!

Answer 1

The way to understand Laplace transforms as applied to differential equations is through the following identity: Let

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

Then

$$\begin{align}\int_0^{\infty} dt \, y'(t) \, e^{-s t} &= \left [ y(t) \, e^{-s t}\right ]_0^{\infty}+ s \int_0^{\infty} dt \, y(t) \, e^{-s t}\\ &= s Y(s) - y(0)\end{align}$$

Similarly, you may show that

$$\int_0^{\infty} dt \, y''(t) \, e^{-s t} = s^2 Y(s) - s y(0) - y'(0)$$

The differential equation then becomes

$$(s^2+4) Y(s) - s y(0) - y'(0) = \int_0^{\pi} dt \, e^{-s t} = \frac{1-e^{-\pi s}}{s}$$

Therefore, taking your initial values,

$$Y(s) = \frac{6 s+7}{s^2+4} + \frac{1-e^{-\pi s}}{s (s^2+4)}$$

Now take the inverse Laplace transform, and you get $y(t)$.

Answer 2

The piecewise function is defined as:

$$f(x)=\begin{cases} 1 & 0 \le x \lt \pi \\ 0 & \pi \le x \lt \infty \end{cases}$$

You can write the Laplace Transform of the piecewise function as:

$$ \int_{0}^{\infty} f(t)~ e^{-st}~dt = \int_{0}^{\pi}~1~e^{-st}~dt + \int_{\pi}^{\infty}~0 ~e^{-st}~dt = \frac{1-e^{-\pi s}}{s} $$

Note: you can also write the piecewise function as a sum of Heaviside Unit Step functions and then take the Laplace Transform using a table of Laplace Transforms.

So,

$$f(t) = 1(u_0(t) - u_{\pi}(t)) + 0(u_{ \pi }(t)) = 1(u(t) - u(t-\pi))$$

This gives:

$$\mathscr{L}~(f(t)) = \mathscr{L}~(u(t) - u(t-\pi)) = \frac{1}{s}-\frac{e^{-\pi s}}{s}$$

Can you take it from here?