Question is Solve using laplace transform i need someone to clarify the notation i dont get the liimits
y'' − y' − 6y = 0 for {0 < t < 2 ; e^t for t > 2}, y(0) = 3, y0(0) = 4
i dont get whats after the brackets
Question is Solve using laplace transform i need someone to clarify the notation i dont get the liimits
y'' − y' − 6y = 0 for {0 < t < 2 ; e^t for t > 2}, y(0) = 3, y0(0) = 4
i dont get whats after the brackets
The right hand side of the equation is a piece wise defined function which is 0 for $\displaystyle 0<t<2$ and $\displaystyle e^t$ for $\displaystyle t>2$, so you could write this as:
$\displaystyle y''-y'-6y=u(t-2) e^t $
where $\displaystyle u $is Heavyside's unit step (note it does not matter that is apprears to now be defined for $\displaystyle t<0$, that part of the solution is not relevant).
CB
I'll give it a try.
$\displaystyle g(t) = \left\{\begin{array}{cc}0,&\mbox{ if } 0 \le t < 2\\e^{t}, & \mbox{ if } t \ge 2\end{array}\right. $
You want to put $\displaystyle g(t) $ in the form $\displaystyle u(t-2)f(t-2) $.
$\displaystyle f(t) = e^{t+2}= e^{2}e^{t} $ because then $\displaystyle f(t-2) = e^{t} $
$\displaystyle y'' - y' -6y = u(t-2)f(t-2) $
$\displaystyle \mathcal{L} \{y''-y'-6y\} = \mathcal{L}\{u(t-2)f(t-2)\} $
$\displaystyle \mathcal{L}\{y'' \} - \mathcal{L} \{y'\} - 6 \mathcal{L}\{y\} = e^{-2s} \frac {e^{2}}{s-1} $
$\displaystyle s^{2} Y(s) -sy(0) -y'(0) - sY(s) + y(0) - 6Y(s) = e^{-2s} \frac {e^{2}}{s-1} $
$\displaystyle s^{2} Y(s) -3s - 4 - sY(s) + 3 - 6Y(s) = e^{-2s} \frac {e^{2}}{s-1} $
$\displaystyle Y(s) (s^{2}-s-6) = e^{-2s} \frac {e^{2}}{s-1} + 3s + 1 $
$\displaystyle Y(s) \big((s+2)(s-3)\big) = e^{-2s} \frac {e^{2}}{s-1} + 3s + 1 $
$\displaystyle Y(s) = e^{-2s} \frac {e^{2}}{(s-1)(s+2)(s-3)} + \frac {3s+1}{(s+2)(s-3)} $
$\displaystyle Y(s) = e^{-2s} \Big(\frac {\text{-}e^{2}}{6} \frac {1}{s-1} + \frac {e^{2}}{15} \frac {1}{s+2} + \frac{e^{2}}{10} \frac{1}{s-3}\Big) + \frac {1}{s+2} + \frac {2}{s-3} $
EDIT: $\displaystyle y(t)= u(t-2) \Big(\frac {\text{-}e^{2}}{6}e^{t-2} + \frac {e^{2}}{15} e^{-2(t-2)} + \frac{e^{2}}{10} e^{3(t-2)} \Big) + e^{-2t} + 2 e^{3t} $
so
$\displaystyle y(t) = \left\{\begin{array}{cc}e^{-2t}+2e^{3t},&\mbox{ if } 0 \le t < 2\\\frac{-e^{t}}{6} + \frac{e^{-2t+6}}{15} + \frac {e^{3t-4}}{10} +e^{-2t} + 2 e^{3t} , & \mbox{ if } t \ge 2\end{array}\right. $
I'm much more confident this time because I get the same answer when I break up the problem into two differential equations with different initial conditions.