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Math Help - Rearranging and Separation of Variables

  1. #1
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    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?
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  2. #2
    Super Member PaulRS's Avatar
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    Divide by: (a^2+s^2)\cdot{<br />
\hat y\left( s \right)<br />
}

    <br />
\left( {a^2  + s^2 } \right) \cdot \hat y\left( s \right) \ne 0<br />

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

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

    And now integrate on both sides
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  3. #3
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    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?
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  4. #4
    Super Member PaulRS's Avatar
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    Quote Originally Posted by ubhik View Post
    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 <br />
{\hat y'\left( s \right)}<br />
=0

    It depends on n, if (s^2+s+n)>0 for all real s we have that: <br />
{\hat y'\left( s \right)}<br />
=0 for all s
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  5. #5
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    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?
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  6. #6
    Super Member PaulRS's Avatar
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    Quote Originally Posted by ubhik View Post
    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
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  7. #7
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    yes...

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

     <br />
(s^2+s+n)\frac{d\hat{y}}{ds}(s)=0<br />

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

    not sure what is in between
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