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Reduction of differntial equation

  1. May 2, 2014 #1
    1. The problem statement, all variables and given/known data

    for this question, i 'm only able reduce it to become x dv/dx = (v^2 + 1) / v ... i have checked it several times, i just couldn't find my mistake...

    2. Relevant equations



    3. The attempt at a solution
     

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  3. May 2, 2014 #2

    SammyS

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    What is [itex]\displaystyle \ \int\frac{v}{v^2+1}\,dv\ ?[/itex]
     
  4. May 2, 2014 #3
    ln (V+1 ) as in my working
     
  5. May 2, 2014 #4

    LCKurtz

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    $$\frac v {v^2+1} \ne \frac v {v(v+1)}$$
     
  6. May 2, 2014 #5
    why cant i do in this way?
     
  7. May 2, 2014 #6

    LCKurtz

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    Because you can't make up your own algebra rules.$$
    \frac v {v(v+1)}= \frac v {v^2+v} \ne \frac v {v^2+1}$$
     
  8. May 2, 2014 #7

    SammyS

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    Take the derivative of ln (V+1 ) .

    It's 1/(V+1) .


    To do that integration, the substitution, u =(v2+1) works very nicely .
     
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