In a chemical reaction $\mathrm{A + 2B \to 3C}$, the concentrations $a(t)$, $b(t)$ and $c(t)$ of the three substances A, B and C measure up to the differential equations $$ \begin{align} \frac{da}{dt} &= -rab^2\tag{1}\\ \frac{db}{dt} &= -2rab^2\tag{2}\\ \frac{dc}{dt} &= -3rab^2\tag{3} \end{align} $$ with $r > 0$ and begin condition $a(0) = 1$ and $b(0) = 2$.
Show that $b(t) - 2a(t) = 0$ .
Here is my solution, but it is not right. Any help would be great.
First equation
$$ \begin{align} \int\frac{da}{a} &= -rb^2\int dt\\ \ln(a) &= -rb^2t + C\\ a(t) &= e^{-rb^2t+ C}\\ a(0) &= 1 &&\to &1 &= e^{0 + C}\\ \ln(1) &= C &&\to &C &= 0\\ a(t) &= e^{-rb^2t} \end{align} $$
Second equation
$$ \begin{align} \frac{db}{dt} &= -2rab^2\\ \int \frac{db}{b^2} &= -2ra\int dt\\ b &= \frac{1}{2rat + C}\\ b(0) &= 2 \quad \to \quad 2 = \frac{1}{C} \quad \to \quad C = \frac{1}{2}\\ b(t) &= \frac{2}{ 2rat + 1} \end{align} $$
But now $b(t) - 2a(t) \ne 0$. Where I am making mistake? Any tip will be enough.
First, we shall show that $b-2a$ is constant, observe $$\frac d{dt}(b-2a)= \frac{db}{dt}-2\frac{da}{dt} $$ And by the system. $$\frac{db}{dt}-2\frac{da}{dt}=-2rab^2+2rab^2=0$$ Hence $b-2a$ is constant. And since $$b(0)-2a(0)=2-2=0$$ We have $b-2a\equiv 0$
Or, you can also solve (“half of”) the equations. Since we have $$\frac{da/dt}{db/dt}=\frac{-rab^2}{-2rab^2}=\frac12$$ Hence, by the chain rule, $$\frac{da}{db}=\frac12\implies a=\frac12b+C_0$$ By the initial condition, we have, $C_0=0$, hence $$b-2a=b-2\frac12b=0$$
Remark: by the second method presented above, you can actually solve for $b(t)$. I leave it to you.