Reversible elementary reaction

I have a question regarding a reversible elementary reaction:


If we have an equilibrium where $$\ce{A<=>[k\_1][k\_{-1}]B}$$ , 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 $$t->\infty$$ 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.


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