1. ## 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

2. Originally Posted by arze
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

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

$
\frac{1}{a} \left(\frac{1}{x} + \frac{1}{a-x}\right) dx = k\, dt.
$