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Simple linear diff.eq.

  1. Apr 11, 2009 #1
    Hey guys.

    I've recently started studying differential equations. There is one thing I don't understand and of which I simply can't find an explanation.

    I'm trying to solve some linear differential equations without using standard solutions.
    Say we have the equation:
    [tex]\frac{dp}{dt}=0.5p - 450[/tex]

    The next step is (according to my book):

    [tex](1) \frac{dp}{p-900}=\frac{1}{2} dt[/tex]

    All of the next steps that lead to the solution are clear to me. They use the chain rule to integrate, exponentiate, and get: [itex]p=900+ce^\frac{t}{2}[/itex].
    But what I don't understand, is why they first write it in the form of eq.(1), and not as, say:
    [tex](2) \frac{dp}{.5p-450}=1 dt[/tex] ?

    Possibly it's a silly question, but nevertheless, please help me out :) .
     
    Last edited: Apr 12, 2009
  2. jcsd
  3. Apr 11, 2009 #2

    Pengwuino

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    Gold Member

    It's equivalent, it's just that using .5 for the right hand side means the coefficient infront of p is simply 1 and the integration is slightly easier. If you used (2), a 2 would simply come out of the left hand side integration as well which makes up for the 1/2 on the right side.

    Also, you do realize there is the 'dt' on the right side correct? It can't just disappear :).
     
  4. Apr 12, 2009 #3
    I thought it should be equivalent, but when I integrate it I get a different solution. I'm trying to integrate it again now to see how I should get the 2 on the left hand side. Thanks for your answer.

    PS: I realized that I forgot the dt's when I had turned of the computer and went to bed :D
     
  5. Apr 12, 2009 #4
    Ah, now I see why the 2 comes in. If you integrate it without the 2, then you forget to compensate for the chain rule factor (or whatever you call it). I don't know if that sentence makes sense, but I get it now :) .
     
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