Chemical kinetics of a reversible reaction
The simplest and most general way to solve this type of scheme is to calculate the effect of an amount x that reacts with initial amounts $A_0$ and $B_0$, thus
$$\ce{A~~~ &lt;=&gt;[k_a][k_b] ~~~B}  \\  A_0-x  ~~~~  B_0 +x $$
which produces the rate equation
$$ \frac{dx}{dt} = k_a(A_0-x)-k_b(B_0-x)$$

the variables can be split to give a straightforward integration instead of having to solve a complicated pair of differential equations.
$$\int _0^x \frac{dx}{k_aA_0-k_bB_0-(k_a+k_b)x} = \int _0^t dt$$

which, after some rearrangement, gives 
$$\ln\left(1-\frac{(k_a+k_b)}{k_aA_0-k_bB_0}\right) = -(k_a+k_b)t$$
and this can be rearranged to
$$x= \left( 1-e^{-(k_a+k_b)t}  \right)\frac{(k_aA_0-k_bB_0)}{k_a+k_b}$$

with the initial condition that at $t=0$, $B_0=\ce{[B]=0 }$ this becomes
$$x= \left( 1-e^{-(k_a+k_b)t}  \right)\frac{k_aA_0}{k_a+k_b}$$
The concentration of A is $A_0-x$ which is

$$ \ce{[A]}_t =A_0 \frac{k_b+k_ae^{-(k_a+k_b)t}}{k_a+k_b}$$

This expression has the correct form at all times. $\ce{[B]}_t$ can be calculated similarly from $B_0+x$.