# Thread: Rearranging and Separation of Variables

1. ## Rearranging and Separation of Variables

I'm following my notes, and I don't understand how to get from:

$(a^2+s^2)\frac{d\hat{y}}{ds}(s)+s\hat{y}(s)=0$

to:

separating the variables and integrating gives

$\int\frac{\hat{y'}(s)}{\hat{y}(s)}ds=-\int\frac{s}{a^2+s^2}ds$

can someone please explain the working that goes in between?

2. Divide by: $(a^2+s^2)\cdot{
\hat y\left( s \right)
}$

$
\left( {a^2 + s^2 } \right) \cdot \hat y\left( s \right) \ne 0
$

You'll get: $
\frac{{\hat y'\left( s \right)}}
{{\hat y\left( s \right)}} + \frac{s}
{{a^2 + s^2 }} = 0
$

So: $
\frac{{\hat y'\left( s \right)}}
{{\hat y\left( s \right)}} = - \frac{s}
{{a^2 + s^2 }}
$

And now integrate on both sides

3. ## Also...

I need to perform a similar operation on the following:

$(s^2+s+n)\frac{d\hat{y}}{ds}(s)=0$

How would I tackle this please?

4. Originally Posted by ubhik
I need to perform a similar operation on the following:

$(s^2+s+n)\frac{d\hat{y}}{ds}(s)=0$

How would I tackle this please?
Ok, for all s we have that either $(s^2+s+n)=0$ or $
{\hat y'\left( s \right)}
=0$

It depends on n, if $(s^2+s+n)>0$ for all real s we have that: $
{\hat y'\left( s \right)}
=0$
for all s

5. ## but...

but if I make one or the other equal to zero, won't that just cancel the whole thing out and make it all zero?

6. Originally Posted by ubhik
but if I make one or the other equal to zero, won't that just cancel the whole thing out and make it all zero?
But it is always equal to 0, right? I mean, the product

7. ## yes...

yes... but in the end I need to obtain something that goes from

$
(s^2+s+n)\frac{d\hat{y}}{ds}(s)=0
$

to $\frac{n!(s-1)^n}{s^{n+1}}$

not sure what is in between
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