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  1. #1
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    Need help

    In a chemical reaction in which a compound X is formed from a compound Y and other substances, the masses of X and Y present at time t are x and y respectively. The sum of the two masses is constant and at any time the rate at which x s increasing is proportional to the product of the two masses at that time. Show that the equation governing the reaction is of the form
    \frac{dx}{dt}=kx(a-x)
    and interpret the constant a. If x=\frac{a}{10} at time t=0, find in terms of k and a the time at which y=\frac{a}{10}

    I have completed the first part where a is the sum of x and y.
    For the next part I know that I have to solve the differential equation and plug the values of t and x.
    \int \frac{1}{kx(a-x)}dx=\int 1dt
    Then I used partial fractions to integrate \int \frac{1}{kx(a-x)}dx
    \int (\frac{\frac{1}{a}}{kax}+\frac{\frac{1}{ka^2}}{a-x})dx=t
    \frac{1}{a}\ln |kax|+\frac{1}{ka^2}\ln |a-x|=t+A
    Now I don't know how to proceed, and I can't get the equation right to find the A.
    Thanks
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  2. #2
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    Quote Originally Posted by arze View Post
    In a chemical reaction in which a compound X is formed from a compound Y and other substances, the masses of X and Y present at time t are x and y respectively. The sum of the two masses is constant and at any time the rate at which x s increasing is proportional to the product of the two masses at that time. Show that the equation governing the reaction is of the form
    \frac{dx}{dt}=kx(a-x)
    and interpret the constant a. If x=\frac{a}{10} at time t=0, find in terms of k and a the time at which y=\frac{a}{10}

    I have completed the first part where a is the sum of x and y.
    For the next part I know that I have to solve the differential equation and plug the values of t and x.
    \int \frac{1}{kx(a-x)}dx=\int 1dt
    Then I used partial fractions to integrate \int \frac{1}{kx(a-x)}dx
    \int (\frac{\frac{1}{a}}{kax}+\frac{\frac{1}{ka^2}}{a-x})dx=t
    \frac{1}{a}\ln |kax|+\frac{1}{ka^2}\ln |a-x|=t+A
    Now I don't know how to proceed, and I can't get the equation right to find the A.
    Thanks
    Not sure how you did your partial fraction decomp. I got

     <br />
\frac{dx}{x(a-x)} = k \,dt<br />

     <br />
\frac{1}{a} \left(\frac{1}{x} + \frac{1}{a-x}\right) dx = k\, dt.<br />
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