# Thread: Differentiation - part of a question

1. ## Differentiation - part of a question

I am in the middle of a question, and following and existing example in my notes, and I have come near the end, but I do not understand a step of the working, can someone please explain, how I get from:

$y(s) = \frac{1}{s^2} - \frac{sy(s)}{s+4}$

to:

$
y(s) = \frac{s+4}{s^2(2s+4)}
$

2. Originally Posted by ubhik
I am in the middle of a question, and following and existing example in my notes, and I have come near the end, but I do not understand a step of the working, can someone please explain, how I get from:

$y(s) = \frac{1}{s^2} - \frac{sy(s)}{s+4}$

to:

$
y(s) = \frac{s+4}{s^2(2s+4)}
$

$y(s) = \frac{1}{s^2} - \frac{sy(s)}{s+4}$

so:

$y(s) \left[1+\frac{s}{s+4}\right]= \frac{1}{s^2}$

hence:

$y(s)= \frac{1}{s^2\left[1+\frac{s}{s+4}\right]}$

Now just tidy up the right hand side.

RonL

3. ## still a bit confused...

so from here

$
y(s)= \frac{1}{s^2\left[1+\frac{s}{s+4}\right]}
$

now I multiply top and bottom with $(s+4)$ right?

so I would get

$
y(s)= \frac{s+4}{s^2\left[1+\frac{s}{s+4}\right](s+4)}
$

Then,

$y(s)= \frac{s+4}{s^2(s+4)+\left[\frac{s^3(s+4)}{s+4}\right]}$

$
y(s)= \frac{s+4}{s^2(s+4)+s^3}
$

but still not sure how to get to

$
y(s) = \frac{s+4}{s^2(2s+4)}
$

4. Originally Posted by ubhik
so from here

$
y(s)= \frac{1}{s^2\left[1+\frac{s}{s+4}\right]}
$

now I multiply top and bottom with $(s+4)$ right?

so I would get

$
y(s)= \frac{s+4}{s^2\left[1+\frac{s}{s+4}\right](s+4)}
$

Then,

$y(s)= \frac{s+4}{s^2(s+4)+\left[\frac{s^3(s+4)}{s+4}\right]}$

$
y(s)= \frac{s+4}{s^2(s+4)+s^3}
$

but still not sure how to get to

$
y(s) = \frac{s+4}{s^2(2s+4)}
$
$s^3 = s^2 * s$. So $s^2(s + 4) + s^3 = s^2(s+4) + s^2(s) = s^2(s + 4 + s) = s^2(2s + 4)$.

5. ## Thank you!!

Greatly appreciated!!