Rearranging and Separation of Variables

**ubhik** · April 30th 2008, 06:25 PM

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?

**PaulRS** · April 30th 2008, 06:30 PM

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

**ubhik** · April 30th 2008, 06:41 PM

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?

**PaulRS** · April 30th 2008, 06:48 PM

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

**ubhik** · April 30th 2008, 06:51 PM

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?

**PaulRS** · April 30th 2008, 06:53 PM

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

**ubhik** · April 30th 2008, 06:59 PM

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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