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.

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.
