The following is a question from Differential Equations with Applications and historical Notes. I have attempted a solution, but I am not sure where I went wrong.
... we expect the rate formation of the compound to be proportional to the number of collisions per unit time, which in turn is jointly proportional to the number of substances that are untransformed. Consider a second order reaction in which $x$ grams of the compound contain $ax$ grams of the first substance and $bx$ grams of the second, where $a+b=1$. If there are $aA$ grams of the first substance and $bB$ grams of the second substance present initially, and $bB$ grams of the second, and if $x=0$ when $t = 0$, find $x$ as a function of time.
This is my attempted solution
$\frac{dx}{dt}=k_1C(A,B)$, where C is a function that gives you the number of collisions based on the amounts of A and B.
$C(A,B) = k_2(A(x)+B(x))$ Where A(x) B(x) are the amounts of A and B remaining depending on how much x is formed
$A(x) = aA -ax$ and $B(x)=(1-a)B-(1-a)x$
Working back up the stack gives:
$C(A,B)=k_2(aA-ax+(1-a)B-(1-a)x)$
$C(A,B)=k_2(aA+B(1-a)-x$
$\frac{dx}{dt}=k_1(k_2(aA+B(1-a)-x))$
And then I solve:
$\frac{dx}{aA+B(1-a)-x}=k_1k_2dt$
$-ln(aA+B(1-a)-x)=k_1k_2t + c$
$aA+B(1-a)-x=\frac{1}{ce^{k_1k_2t}}$
$x = aA+B(1-A)-\frac{1}{ce^{k_1k_2t}}$
I don't really know what to do from here, because I am not confident that I did this right....Do the steps up to here seem correct? When I solve for initial conditions I do n ot get the same answer as the back of the book.
Well, first, having both $k_1$ and $k_2$ is redundant. Second, I believe that $C(A,B)$ should be $k(A(x)*B(x))$ rather than $k(A(x)+B(x))$. See, for instance, http://www.mathwords.com/jk/joint_variation.htm . "$a$ is jointly proportional to $b$ and $c$" means that if either $b$ or $c$ is doubled, then $a$ is doubled, and if both are doubled, then $a$ is quadrupled; if $b$ is treated as a constant, then $a$ is proportional to $c$, and if $c$ is treated as a constant, then $a$ is proportional to $b$.