Using Laplace transform solve the following differential equation :
$$ y'+y = 2 + \delta(t-4)$$, $$y(0)=0$$
Express the solution $y(t)$ as a piecewise function about $t=4$ and tell us what happens to the graph of it at $t=4$.
My try :
Taking Laplace transform on the both sides :
$$ \big( s \cdot Y(s) – y(0) \big) + Y(s) = \frac{2}{s} + e^{-4s} \\
\implies Y(s) = \frac{2}{s(s+1)} + \frac{e^{-4s}}{(s+1)} \\
\implies Y(s) = 2\Big(\frac{1}{s}-\frac{1}{s+1}\Big) + \frac{e^{-4s}}{(s+1)}
$$
Now I'm stuck here because I dont know how to take the inverse Laplace transform of $\frac{e^{-4s}}{(s+1)}$ . I haven't ever dealt with something involving exponential function. Can I get some help here please?
Best Answer
You can use: $$\mathcal {L}^{-1}\{e^{-cs}F(s)\}=u(t-c)f(t-c)$$ You have a Laplace transform table here