Reversible elementary reaction I have a question regarding a reversible elementary reaction:

If we have an equilibrium where $$\ce{A&lt;=&gt;[k_1][k_{-1}]B}$$ , then $$-d[A]/dt= k_1[A]-k_{-1}[B]$$.

If we take $$[B]=[B]_0 + ([A]_0 - [A])$$, the equation above becomes $$d[A]/dt= -k_1[A]+k_{-1}([B]_0+([A]_0 -[A])$$ 

So far, I've taken the integrating factor to be $$u=\mathrm{e}^{(k_1+k_{-1})t}$$

and emerged with the following equation:

$$[A]=\frac{k_{-1}}{k_1+k_{-1}}([B]_0+[A]_0)(1-\mathrm{e}^{-t(k_1+k_{-1})}) + C \mathrm{e}^{-t(k_1+k_{-1})}$$

I have tried to put $$[A](t=0) = [A]_0$$ and the solution appears to be $$C=[A]_0$$

Are there further approximations required?

My lecturer has produced the final equation as 

$$[A]=\frac{k_{-1}+k_1\mathrm{e}^{-t(k_1+k_{-1})}}{k_1+k_{-1}}([B]_0+[A]_0)$$

When we take $$t-&gt;\infty$$ our equations match, however.
I went through the derivation myself and I can reproduce the equation you got. But after using $C = [\ce{A}]_0$ and some rearrangements I get the following formula

\begin{equation}
\ce{[A]} = \frac{k_{-1}}{k_{1} + k_{-1}} \left( \ce{[A]_{0}} + \ce{[B]_{0}} \right) + \frac{\mathrm{e}^{-(k_{1} + k_{-1})t} }{k_{1} + k_{-1}} \left( \ce{[A]_{0}} k_{1} - \ce{[B]_{0}} k_{-1} \right)
\end{equation}

and not the one your professor got. But your professor's formula might be incorrect because for $t=0$ his formula yields the incorrect result $\ce{[A]}_{t = 0} = \ce{[A]_{0}} + \ce{[B]_{0}}$ instead of the correct result $\ce{[A]}_{t =  0} = \ce{[A]_{0}}$, so it doesn't satisfy the given boundary conditions.