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Getting started with differential equations

  1. Feb 23, 2005 #1
    Folks, I'm just getting started with differential equations and I need some help.

    [tex]\frac{dp}{dt}=0.5p-450 \qquad (1)[/tex]

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

    [tex]\frac{d}{dt}\ln \left| p-900 \right| = \frac{1}{2} \qquad (3)[/tex]

    I'm stuck in [tex](3)[/tex]. I know

    [tex]\frac{d}{dt}\ln \left| x \right| = \frac{1}{x}[/tex]

    but I still don't understand how that is obtained.

    Any help is highly appreciated.
     
    Last edited: Feb 23, 2005
  2. jcsd
  3. Feb 23, 2005 #2

    arildno

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    What do you mean?
    [tex]\frac{d}{dt}ln|x(t)|=\frac{dx}{dt}\frac{1}{x}[/tex]
     
  4. Feb 23, 2005 #3
    I've just edited my previous post. There was a typo.

    Well, the fact that in general

    [tex]\frac{d}{dt}\ln \left| x \right| = \frac{1}{x}[/tex]

    should give some insight into the transition from (2) to (3), but I still don't get it. That's what I mean.

    Thanks
     
    Last edited: Feb 23, 2005
  5. Feb 23, 2005 #4

    arildno

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    First off, this is WRONG!!!
    It should be:
    [tex]\frac{d}{dt}\ln \left| x \right| = \frac{1}{x}\frac{dx}{dt}[/tex]

    Do you now see the connection between (2) and (3)?
     
  6. Feb 23, 2005 #5
    Ooops. I'm sorry (it was a typo). I mean

    [tex]\frac{d}{dx}\ln \left| x \right|=\frac{1}{x}[/tex]

    I still don't get the connection between (2) and (3). Could you please clarify that?
     
  7. Feb 23, 2005 #6

    arildno

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    Note that, by the chain rule we have :
    [tex]\frac{d}{dt}ln|p(t)-900|=\frac{1}{p-900}\frac{dp}{dt}=\frac{\frac{dp}{dt}}{p-900}[/tex]
    Does that help you?
     
  8. Feb 23, 2005 #7

    dextercioby

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    Have you been taught the METHOD OF SEPARATION OF VARIABLES...?I.e.Working with differentials,just as if they were numbers which could be multiplied & divided with...?

    Daniel.
     
  9. Feb 23, 2005 #8
    So, my problem is with the chain rule step. Arildno, I don't see why

    [tex]\frac{d}{dt}ln|p(t)-900|=\frac{1}{p-900}\frac{dp}{dt}[/tex]

    although I can apply it (for example):

    [tex]\frac{d}{dx}2(4x+1)^2=2\cdot 2 \cdot (4x+1)\cdot 4=64x+16[/tex]

    Could please you explain me how that works?

    Daniel, I've read about separation of variables. That's going to be useful once I'm done with (3) and move on.
     
  10. Feb 23, 2005 #9

    arildno

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    Well, the chain rule states, letting x=p-900:
    [tex]\frac{d}{dt}ln|p-900|=\frac{d}{dt}ln|x|=(\frac{d}{dx}ln|x|)\frac{dx}{dp}\frac{dp}{dt},\frac{d}{dx}ln|x|=\frac{1}{x},\frac{dx}{dp}=1[/tex]

    were you interested in a more detailed explanation behind the chain rule?
     
  11. Feb 23, 2005 #10

    dextercioby

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    What is [tex] \frac{d\ln u(t)}{dt} [/tex]...?You said u knew how to apply the chain rule...


    Daniel.
     
  12. Feb 23, 2005 #11
    I can now see the way to get there:

    [tex]\frac{dp}{dt}=0.5p-450 \qquad (1)[/tex]

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

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

    [tex]\frac{dp}{dt} \left[ \frac{d}{dp} \ln \left| p-900 \right| \right] =\frac{1}{2} [/tex]

    [tex]\frac{d}{dt}\ln \left| p-900 \right| = \frac{1}{2} \qquad (3)[/tex]

    Thank you
     
  13. Feb 23, 2005 #12

    dextercioby

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    Okay,how about posting the sollution to this ODE...?

    Daniel.
     
  14. Feb 23, 2005 #13
    [tex]\frac{dp}{dt}=0.5p-450 [/tex]

    [tex]\frac{dp/dt}{p-900}=\frac{1}{2} \qquad \mbox{if }p \neq 900[/tex]

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

    [tex]\frac{dp}{dt} \left[ \frac{d}{dp} \ln \left| p-900 \right| \right] =\frac{1}{2} [/tex]

    [tex]\frac{d}{dt}\ln \left| p-900 \right| = \frac{1}{2}[/tex]

    [tex] \int \left[ \frac{d}{dt}\ln \left| p-900 \right| \right] dt = \int \frac{1}{2} dt[/tex]

    [tex] \ln \left| p-900 \right| = \frac{t}{2} + \mathrm{C} [/tex]

    [tex] \left| p-900 \right| = e^{\frac{t}{2} + \mathrm{C}} [/tex]

    [tex] \left| p-900 \right| = e^{\mathrm{C}}e^{\frac{t}{2}} [/tex]

    [tex] p-900 = \pm e^{\mathrm{C}}e^{\frac{t}{2}} [/tex]

    [tex] p(t) = 900 + \mathrm{K}e^{\frac{t}{2}} \quad \mbox{where } \mathrm{K}=\pm e^{\mathrm{C}} \mbox{ is an arbitrary constant.}[/tex]
     
    Last edited: Feb 23, 2005
  15. Feb 23, 2005 #14

    arildno

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    That is correct.
     
  16. Feb 23, 2005 #15

    dextercioby

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    Look at the bright side.Once u'll learn separation of variables,u'll be able to shortcut many calculations...:wink:

    Daniel.
     
  17. Feb 23, 2005 #16
    Yes, indeed. :tongue2:

    [tex]\frac{dp}{dt}=0.5p-450 [/tex]

    [tex]\frac{dp/dt}{p-900}=\frac{1}{2} \qquad \mbox{if }p \neq 900[/tex]

    [tex]\int \frac{dp}{p-900}=\int \frac{dt}{2}[/tex]

    [tex] \ln \left| p-900 \right| = \frac{t}{2} + \mathrm{C} [/tex]

    [tex] \left| p-900 \right| = e^{\frac{t}{2} + \mathrm{C}} [/tex]

    [tex] \left| p-900 \right| = e^{\mathrm{C}}e^{\frac{t}{2}} [/tex]

    [tex] p-900 = \pm e^{\mathrm{C}}e^{\frac{t}{2}} [/tex]

    [tex] p(t) = 900 + \mathrm{K}e^{\frac{t}{2}} \quad \mbox{where } \mathrm{K}=\pm e^{\mathrm{C}} \mbox{ is an arbitrary constant.}[/tex]
     
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