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    I'm really stuck with a practical differential equation question!

    I can't work out how to do the last part (see picture) where I have to find a limiting value for x?

    Anyone have any ideas?

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    Rearrange to get x in terms of t.
    Then rearrange the function of t to work out the limit.
    Tell me if you want more details.
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    (Original post by morgan8002)
    Rearrange to get x in terms of t.
    Then rearrange the function of t to work out the limit.
    What do you mean rearrange the function of t to get the limit? I thought you needed the value of constant k...which I don't see how I can work out with only x=0 at t=0. :confused:
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    (Original post by Mr_Pickles3)
    What do you mean rearrange the function of t to get the limit? I thought you needed the value of constant k...which I don't see how I can work out with only x=0 at t=0. :confused:
    Use (0, 0) to find the value of c.
    The value of k becomes irrelevant later on.
    You need to rearrange to get x in terms of t. There is one more step after that before the limit becomes apparent.
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    (Original post by morgan8002)
    Use (0, 0) to find the value of c.
    The value of k becomes irrelevant later on.
    You need to rearrange to get x in terms of t. There is one more step after that before the limit becomes apparent.
    I managed to rearrange to get that and have found that c=0, but I'm still left with an equation with no values to substitute in that will work....

    Sorry but I'm still confused :dontknow:Name:  1421354277058.jpg
Views: 75
Size:  28.8 KBName:  1421354277058.jpg
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    (Original post by Mr_Pickles3)
    I managed to rearrange to get that and have found that c=0, but I'm still left with an equation with no values to substitute in that will work....

    Sorry but I'm still confused :dontknow:Name:  1421354277058.jpg
Views: 75
Size:  28.8 KBName:  1421354277058.jpg
Views: 75
Size:  28.8 KB

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    That's roughly correct, but I got the negative of that and can't find a mistake. Check your working.
    Now divide the top and bottom by e2kt, as you would to find the limits of any rational function.
    edit: look at lines 2-3, also lines 5-6
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    (Original post by morgan8002)
    That's roughly correct, but I got the negative of that and can't find a mistake. Check your working.
    Now divide the top and bottom by e2kt, as you would to find the limits of any rational function.
    edit: look at lines 2-3, also lines 5-6
    Oh yeah...whoops

    Okay, I'm still very stuck....I think I'll just have to give up for now :mad:

    Thanks for all your help though
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    (Original post by Mr_Pickles3)
    Oh yeah...whoops

    Okay, I'm still very stuck....I think I'll just have to give up for now :mad:

    Thanks for all your help though
    You nearly got it out. Maybe look at rational functions in FP1 or limits in FP2/FP3 because the methods are similar from now.
    x = \frac{e^{2kt} - 1}{4e^{2kt} - 2}

    x = \frac{1 - e^{-2kt}}{4 - 2e^{-2kt}}

    \therefore t \rightarrow \infty \implies x \rightarrow \frac{1}{4}
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    (Original post by morgan8002)
    You nearly got it out. Maybe look at rational functions in FP1 or limits in FP2/FP3 because the methods are similar from now.
    x = \frac{e^{2kt} - 1}{4e^{2kt} - 2}

    x = \frac{1 - e^{-2kt}}{4 - 2e^{-2kt}}

    \therefore t \rightarrow \infty \implies x \rightarrow \frac{1}{4}

    Ah, I'm not doing anything above C4 which might explain why I've been struggling then
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    (Original post by Mr_Pickles3)
    Ah, I'm not doing anything above C4 which might explain why I've been struggling then
    It's not tricky and the method is explained in FP1. I suggest you borrow an FP1 textbook from the library or something and read up on limits of rational functions.


    The idea is that you divide by the variable enough times so that all you have left on the top or the bottom are a number and then a sum of negative powers.



    For example, find the limit as x tends to infinity: \frac{3x^2 + 2x + 4}{2x^2 -x +3}
    You divide by x^2 to get \frac{3 + 2x^{-1} +4x^{-2}}{2 - x^{-1} +3x^{-2}}

    As x tends to infinity, all the terms except for the first on the top and bottom tend to 0.
    So limit = \frac{3}{2}
    I applied the same method to your question.
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    (Original post by Mr_Pickles3)
    Ah, I'm not doing anything above C4 which might explain why I've been struggling then
    have some practice here with full worked solutions

    http://madasmaths.com/archive/maths_..._modelling.pdf

    long download time.
    I hope it helps
 
 
 
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