Results 1 to 2 of 2

Thread: Need help

  1. #1
    Senior Member
    Joined
    Jul 2009
    From
    Singapore
    Posts
    338

    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
    $\displaystyle \frac{dx}{dt}=kx(a-x)$
    and interpret the constant a. If $\displaystyle x=\frac{a}{10}$ at time t=0, find in terms of k and a the time at which $\displaystyle y=\frac{a}{10}$

    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.
    $\displaystyle \int \frac{1}{kx(a-x)}dx=\int 1dt$
    Then I used partial fractions to integrate $\displaystyle \int \frac{1}{kx(a-x)}dx$
    $\displaystyle \int (\frac{\frac{1}{a}}{kax}+\frac{\frac{1}{ka^2}}{a-x})dx=t$
    $\displaystyle \frac{1}{a}\ln |kax|+\frac{1}{ka^2}\ln |a-x|=t+A$
    Now I don't know how to proceed, and I can't get the equation right to find the A.
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    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
    $\displaystyle \frac{dx}{dt}=kx(a-x)$
    and interpret the constant a. If $\displaystyle x=\frac{a}{10}$ at time t=0, find in terms of k and a the time at which $\displaystyle y=\frac{a}{10}$

    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.
    $\displaystyle \int \frac{1}{kx(a-x)}dx=\int 1dt$
    Then I used partial fractions to integrate $\displaystyle \int \frac{1}{kx(a-x)}dx$
    $\displaystyle \int (\frac{\frac{1}{a}}{kax}+\frac{\frac{1}{ka^2}}{a-x})dx=t$
    $\displaystyle \frac{1}{a}\ln |kax|+\frac{1}{ka^2}\ln |a-x|=t+A$
    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

    $\displaystyle
    \frac{dx}{x(a-x)} = k \,dt
    $

    $\displaystyle
    \frac{1}{a} \left(\frac{1}{x} + \frac{1}{a-x}\right) dx = k\, dt.
    $
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum