Heaviside function

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• Nov 4th 2009, 03:34 AM
macduff
Heaviside function
I am having a bit of trouble with heavy side functions. One of the questions from a past exam paper I am looking at asks:

Use Laplace transforms to solve the initial value problem
$y"+4y=H(t-1)$
subject to initial conditions
$y(0)=-1, y'(0)=-1$
Here H is the Heaviside (unit step function). I am totally confused on how to go about this. If anyone could please show me step by step I would greatly appreciate it because my lecturers notes on this particular question don't really explain much. :)
• Nov 4th 2009, 04:38 AM
chisigma
The equation written in terms of LT is...

$s^{2}\cdot Y(s) -s\cdot y(0) - y^{'} (0) +4\cdot Y(s) = \frac{e^{-s}}{s}$ (1)

... and from (1) we obtain $Y(s)$ as...

$Y(s) = \frac{s\cdot y(0) + y^{'} (0)}{s^{2} + 4} + \frac{e^{-s}}{s\cdot (s^{2}+4)}$ (2)

The solution $y(t)$ is the inverse LT of (2)...

$y(t)=\mathcal{L}^{-1} \{Y(s)\}= y(0)\cdot \cos 2t + y^{'} (0) \cdot \sin 2t + \frac{1}{4}\cdot \{1 - \cos 2\cdot(t-1)\}\cdot H(t-1)$ (3)

Kind regards

$\chi$ $\sigma$
• Nov 4th 2009, 05:52 AM
CaptainBlack
Quote:

Originally Posted by macduff
I am having a bit of trouble with heavy side functions. One of the questions from a past exam paper I am looking at asks:

Use Laplace transforms to solve the initial value problem
$y"+4y=H(t-1)$
subject to initial conditions
$y(0)=-1, y'(0)=-1$
Here H is the Heaviside (unit step function). I am totally confused on how to go about this. If anyone could please show me step by step I would greatly appreciate it because my lecturers notes on this particular question don't really explain much. :)

You can solve this using Laplace transforms as chisigma has explained. You can also solve this in the time domain by solving the IVP:

$y''+4y=0$, $y(0)=-1, y'(0)=-1$ and using the solution when $t \le 1$, then for $t\ge1$ solving the IVP:

$y''+4y=1$

with initial conditions at $x=1$ determined from the other solution at this point.

CB