1. ## 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.

2. 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

$\chi$ $\sigma$

3. 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