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**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

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

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