I am having trouble figuring out what exactly to do for this question. Given the initial conditions y(0)=1 and the equation y'-2y=4-3u(t-2) where u is the heaviside unit step function. I took the Laplace of both sides and got: sy(s)-1+2y(s)= 4/s - L{3u(t-2)}. I know that the left side simplifies to: (s+2)y(s) but I am not sure how to simplify L{3u(t-2)}.
$\endgroup$ 12 Answers
$\begingroup$Can work it out using the following idea (split the integral):
$L\{u(t-2)\} = \int_0^\infty{u(t-2)e^{-st}dt} = \int_0^2{u(t-2)e^{-st}dt} + \int_2^\infty{u(t-2)e^{-st}dt} = \int_2^\infty{e^{-st}dt} = \ldots = \frac{e^{-2s}}{s}$
$\endgroup$ $\begingroup$Since $u(t-2)=0$ for $t<2$, but $1$ for $t>2$, $$\mathcal{L}(3u(t-2)) = \int_0^\infty e^{-st}3 \cdot u(t-2) \; dt$$ $$=3\int_2^\infty e^{-st} \; dt$$ $$=-\left.\frac{e^{-st}}{s}\right|_2^\infty$$ $$=\frac{e^{-2s}}{s}$$
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