Solve differnetial eqation laplace transforms

**zangestu888** · July 13th 2009, 09:31 PM

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

**CaptainBlack** · July 13th 2009, 10:16 PM

Originally Posted by zangestu888

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 $0<t<2$ and $e^t$ for $t>2$ , so you could write this as:

$y''-y'-6y=u(t-2) e^t$

where $u$ is Heavyside's unit step (note it does not matter that is apprears to now be defined for $t<0$ , that part of the solution is not relevant).

CB

**Random Variable** · July 14th 2009, 03:00 PM

I'll give it a try.

$g(t) = \left\{\begin{array}{cc}0,&\mbox{ if } 0 \le t < 2\\e^{t}, & \mbox{ if } t \ge 2\end{array}\right.$

You want to put $g(t)$ in the form $u(t-2)f(t-2)$ .

$f(t) = e^{t+2}= e^{2}e^{t}$ because then $f(t-2) = e^{t}$

$y'' - y' -6y = u(t-2)f(t-2)$

$\mathcal{L} \{y''-y'-6y\} = \mathcal{L}\{u(t-2)f(t-2)\}$

$\mathcal{L}\{y'' \} - \mathcal{L} \{y'\} - 6 \mathcal{L}\{y\} = e^{-2s} \frac {e^{2}}{s-1}$

$s^{2} Y(s) -sy(0) -y'(0) - sY(s) + y(0) - 6Y(s) = e^{-2s} \frac {e^{2}}{s-1}$

$s^{2} Y(s) -3s - 4 - sY(s) + 3 - 6Y(s) = e^{-2s} \frac {e^{2}}{s-1}$

$Y(s) (s^{2}-s-6) = e^{-2s} \frac {e^{2}}{s-1} + 3s + 1$

$Y(s) \big((s+2)(s-3)\big) = e^{-2s} \frac {e^{2}}{s-1} + 3s + 1$

$Y(s) = e^{-2s} \frac {e^{2}}{(s-1)(s+2)(s-3)} + \frac {3s+1}{(s+2)(s-3)}$

$Y(s) = e^{-2s} \Big(\frac {\text{-}e^{2}}{6} \frac {1}{s-1} + \frac {e^{2}}{15} \frac {1}{s+2} + \frac{e^{2}}{10} \frac{1}{s-3}\Big) + \frac {1}{s+2} + \frac {2}{s-3}$

EDIT: $y(t)= u(t-2) \Big(\frac {\text{-}e^{2}}{6}e^{t-2} + \frac {e^{2}}{15} e^{-2(t-2)} + \frac{e^{2}}{10} e^{3(t-2)} \Big) + e^{-2t} + 2 e^{3t}$

so

$y(t) = \left\{\begin{array}{cc}e^{-2t}+2e^{3t},&\mbox{ if } 0 \le t < 2\\\frac{-e^{t}}{6} + \frac{e^{-2t+6}}{15} + \frac {e^{3t-4}}{10} +e^{-2t} + 2 e^{3t} , & \mbox{ if } t \ge 2\end{array}\right.$

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.

**zangestu888** · July 14th 2009, 07:36 PM

thanks!! i did it also from scratch i got the same answer! except took me a while with stupied mistakes

**Random Variable** · July 14th 2009, 09:05 PM

Originally Posted by zangestu888

thanks!! i did it also from scratch i got the same answer! except took me a while with stupied mistakes

I edited my post above.

Thread: Solve differnetial eqation laplace transforms

LinkBack

Thread Tools

Display

Solve differnetial eqation laplace transforms

Similar Math Help Forum Discussions

"Solve y'' - 3y' + 4y = 0; y(0) = 1, y'(0) = 5" (using Laplace transforms)

Laplace Transforms to solve Initial-Value problem

Laplace Transforms to solve First Order ODE's.

Laplace Transforms

laplace transforms to solve IVPs

Search Tags