Laplace Transform with Piecewise...

Question

I've been having a hard time with Laplace. Did I find $Y(s)$ correctly? $$ y'+2y = f(t), \quad f(t) = \begin{cases}0,\;&0\le t<3,\\ 4,\;&3\le t. \end{cases} $$ where $y(0) = 1$

the attempt:

$$ g(t) = 0, h(t) =4, a=3 $$

$$ f(t) = 4u(t-3) $$ $$ y'+2y=4u(t-3) $$ $$ sY(s)+sY(0)+2Y(s) = \frac 4{s}e^{-3s} $$ isolate $Y(s)$ to get $$ Y(s) = \frac {4e^{-3s}}{s(s+2)}-\frac 1{s+2} $$

however, I am unsure if I did the steps correctly. I'm also new here so I really have a hard time typing this all out :(

Answer 1

You're doing just fine with the typing! You'll get better as you get more used to it. You can check your work in the Wolfram Dev Platform by executing the command Solve[LaplaceTransform[y'[t]+2y[t]==4 UnitStep[t-3],t,s],LaplaceTransform[y[t],t,s]], which yields $$\mathcal{L}\{y\}=\dfrac{e^{-3s}(4+e^{3s}\,s\,y(0))}{s(s+2)}.$$ You can then check your fraction decomposition by using the Apart function to get $$\mathcal{L}\{y\}=\dfrac{4e^{-3s}}{s(s+2)}+\dfrac{y(0)}{s+2},$$ which differs from your answer by a sign, I believe.

Don't forget that $\mathcal{L}\{y'(t)\}=s\mathcal{L}\{y\}-y(0)$.