Differential equations for chemical reaction $\mathrm{A + 2B \to 3C}$ 
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.
 A: 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.
A: Notice that, by your equations,
$\dfrac{d(2a - b)}{dt} = 2\dfrac{da}{dt} - \dfrac{db}{dt} = -2rab^2 -(-2rab^2) = 0; \tag 1$
hence $2a(t) - b(t)$ is constant.  Now
$2a(0) - b(0) = 2(1) - 2 = 0; \tag 2$
the desired result follows.
A: The error is in your solution of the second equation: you forgot that the concentration $a(t)$ is a function of $t$, you are not allowed to pull it out before the integration sign.
