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Thread: Some differential equation problems.

  1. #1
    Sep 2012

    Some differential equation problems.

    Hello everyone,

    Could someone please help me with this problem. It is related to chemical reactions but the behaviour can be described by differential equations. I've simplified the model to the equations you see below.

    Lets say there are have two reactions which can be described by the equations:

    1: dy/dt = -a*y
    2: dy/dt = -b*y

    where a and b are constants.

    Integration gives:

    1: y(t) = y(0) * exp(-a*t)
    2: y(t) = y(0) * exp(-b*t)

    Now we perform an experiment where both systems are present so the observed behaviour can be described by:

    3: dy/dt = -a*y -b*y

    and integrated:

    3: y(t) = y(0) * exp(-(a + b)*t)

    Now what I want to know is the following:

    After the experiment I obtain results described by 3. I know the background reaction is described by 2. However I am only interested in the exact results of reaction 1. How can I extract the exact information of 1 out of 3 as if 2 was never present. In othe words I need to correct 3 to obtain the actual behaviour I'm interested in.

    I thought it would be simple substraction/addition for instance in the simplest approach one could say to just 'correct' reaction 3 like:

    4: dy/dt = -a*y - b*y + b*y

    however this equation (when plotted) does not produce the same outcome as 1 because it does not account for the change in y caused by reaction 2.

    So it looks like that: dy/dt = -a*y - b*y + b*y is not the same as dy/dt = -a*y

    And that is where my mathematical knowledge ends
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  2. #2
    Member kalyanram's Avatar
    Jun 2008
    Uppsala, Sweden

    Re: Some differential equation problems.

    Did you try comparing the factor y(0) that you are multiplying with in both cases. The best way to compare them is obtain the y(t) = y(0) * e^{(-(a + b)*t)} at t=0 and that of y(t) = y(0) * e^{(-a*t)} at t=0 and see if the value of the graph at this point tally.

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