Question

PLEASE HELP!!!!!!!!!!!!!!
Consider the piecewise continuous function
\[f(t)=\left\{\begin{matrix} t-3, &t<2 \\ 1, & t\geq 2 \end{matrix}\right.\]
Rewrite f(t) in terms of unit step functions.

Solve the IVP \[y''-4y'+4y=f(t), y(0)=-1, y'(0)=-2\]

Answer 1

Make use of the Laplace transform. You may want to consider writing the given \(f(t)\) in terms of the Heaviside step function.

Answer 2

^though doing so isn't absolutely necessary. \(f\) is simple enough to have each piece's transform computed quite easily from the integral definition.

Answer 3

\[(u-t)U_{2}(t)+U_{2}(t)\]
Is that how you re-write the piecewise

Answer 4

I'm not sure I understand your notation. What's the distinction between \(u\) and \(U\)?

Answer 5

I denote the step function as
\[u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t<c\end{cases}\]
You should have
\[f(t)=(t-3)u(-t+2)+u(t-2)\]

Answer 6

for some reason that's how my professor showed us how to do but when i read the book, it's the same as you have it. so am a little bit confused

Answer 7

Your prof might be using a slightly different notation, where \(u_c(t)=u(t-c)\) mean the same thing.

Answer 8

Oh okay. that's now makes sense because it has been confusing me!

Answer 9

\[f(t)=(t-3)u(-t+2)+u(t-2)\]
\[\begin{align*}
\mathcal{L}\{f(t)\}&=\int_0^\infty (t-3)u(-t+2)e^{-st}~dt+\int_0^\infty u(t-2)e^{-st}~dt\\\\
&=\int_0^2 (t-3)e^{-st}~dt+\int_2^\infty e^{-st}~dt\\\\
&=\left[-\frac{1}{s}(t-3)e^{-st}\right]_0^2+\frac{1}{s}\int_0^2 e^{-st}~dt+\int_2^\infty e^{-st}~dt\\\\
&=\frac{e^{-2s}-3}{s}+\frac{1-e^{-2s}}{s^2}+\frac{e^{-2s}}{s}
\end{align*}\]
and keep going from there.