Correct Rate Equation for a Reversible Reaction
If all you want to find is the final concentration of all the species at equilibrium than you don't need to solve these differential equations.
To solve this types of questions easily in chemistry we define a quantity $K_{eq}$. Where $K_{eq}=\frac{K_f}{K_b}$ and $K_{eq}=\frac{\text{product of concentrations of products at equilibrium to the power of their stoichiometric coefficient}}{\text{product of concentrations of reactants at equilibrium to the power of their stoichiometric coefficient}}$.
So for your case we can set the equations as such:
$$
\begin{align}
\begin{array}{cccccc}
&amp; \ce{A} &amp; \ce{+} &amp; \ce{B} &amp; \ce{&lt;=&gt;} &amp; \ce{C}\\
\text{Initial} &amp; \pu{1 M} &amp; &amp; \pu{1 M} &amp; &amp; \pu{0 M}\\
\text{Final} &amp; (1-x)\pu{M} &amp; &amp; (1-x)\pu{M} &amp; &amp; x\pu{M}
\end{array}
\end{align}
$$
$$
K_{eq}=\frac{K_f}{K_b}=\frac{[\ce{C}]}{[\ce{A}][\ce{B}]}\\
2=\frac{x}{(1-x)^2}\\
2x^2-5x+2=0\\
(2x^2-4x)+(-x+2)=0\\
(2x-1)(x-2)=0\\
x=\{\pu{0.5M}, \pu{2M}\}\\
$$
But as $[\ce{B}]\leq \pu{1M}$, $x$ has to be equal to $\pu{0.5 M}$.
Which gives, $[\ce{A}]=[\ce{B}]=[\ce{C}]=\pu{0.5 M}$.
This also shows that your solving of differential equations was also correct!