# Laplace transform

1. May 12, 2009

### Ry122

http://users.on.net/~rohanlal/laplacetrans5.jpg [Broken]
In the third line, what is S+4 the laplace transform of?

Last edited by a moderator: May 4, 2017
2. May 12, 2009

### gabbagabbahey

It doesn't matter. Just isolate Y(s) by dividing both sides by (s+4). Then use partial fractions to compute the inverse Laplace Transform of the RHS.

3. May 13, 2009

### Ry122

it does matter because this is part of the solution to the problem and i need to understand how its done.

4. May 13, 2009

### HallsofIvy

Staff Emeritus
No, "the function whose Laplace transform is s+ 4" is not part of the solution, the function whose Laplace transform is 1/(s+ 4) is.

As Gabbagabbahey said, you need to divide both sides by s+ 4 to get
$$Y(s)= \frac{17}{(s^2+1)(s+4)}+ \frac{1}{s+4}$$
Partial fractions will give something of the form
$$Y(s)= \frac{As}{s^2+1}+ \frac{B}{s^2+1}+ \frac{C}{s+4}$$
and then you can look up the inverse transform of each of those.

5. May 13, 2009

### Ry122

so is the laplace transform of 4y = 1/s+4? then how does it end up being a factor of y(s)?

6. May 13, 2009

### gabbagabbahey

I'm not exactly sure what you are trying to ask here. If you are asking whether $4Y(s)=\frac{1}{s+4}$, then the answer is no.

You have correctly gotten to this point:

$$(s+4)Y(s)=\frac{17}{s^2+1}+1$$

However, for this to be useful you need to divide both sides of the equation by $s+4$ in order to isolate $Y(s)$. The reason why simply taking the inverse Laplace transform of both sides in the current form is useless is because $\mathcal{L}^{-1}[(s+4)Y(s)]\neq(\mathcal{L}^{-1}[(s+4)])\cdot(\mathcal{L}^{-1}[Y(s)])$

But if you divide by $s+4$ first, you get

$$Y(s)=\frac{17}{(s^2+1)(s+4)}+\frac{1}{s+4} \implies y(t)=\mathcal{L}^{-1}[Y(s)]=\mathcal{L}^{-1}[\frac{17}{(s^2+1)(s+4)}+\frac{1}{s+4}]$$

which allows you to find $y(t)$.

7. May 13, 2009

### Ry122

The working i attached wasn't carried out by me. I havent gotten as far as
$$(s+4)Y(s)=\frac{17}{s^2+1}+1$$

someone else did. i want to know how they got there.
im unsure as to how they got y(s)[s+4]
i also dont know what the laplace transform of 4y is.

8. May 13, 2009

### gabbagabbahey

Oh! Now I see what you were asking.

The workings in your attached image should be fairly clear: $\mathcal{L}[4y(t)]=4\mathcal{L}[y(t)]=4Y(s)$ and $\mathcal{L}[y'(t)]=sY(s)-y(0)=sY(s)-1$. The reasons for this should be in your textbook/notes, and if you are not comfortable with such basic Laplace transforms you need to study your text/notes and perhaps get a tutor.