Chemical kinetics with the reaction of tert-butyl bromide with azide ion The reaction of tert-butyl bromide with azide ion in aqueous solution
  is proposed to proceed by the following mechanism:
  
  $\ce{(CH3)3CBr(aq) &lt;=&gt;(CH3)3C+(aq) + Br–(aq)}$
  
  $\ce{(CH3)3C+(aq) + N3–(aq) \to (CH3)3CN3(aq)}$
  
  Assuming that $\ce{(CH3)3C+ (aq)}$ achieves a steady-state concentration, but making no further assumptions about the relative magnitudes of the three rate constants,
  what is the rate law for this reaction, where $k_1$ is the forward
  reaction rate of the first equation, $k_{-1}$ is the reverse reaction
  rate of the first equation, and $k_2$ is the forward reaction rate of
  the second equation?
  
  (A) $\ce{k_1[(CH3)3CBr]}$
  
  (B) $\ce{k_2[(CH3)3CBr][N3-]}$
  
  (C) $\ce{\displaystyle\frac{k_1k_2[(CH3)3CBr][N3-]}{k_{-1}[Br-]}}$
  
  (D) $\ce{\displaystyle\frac{k_1k_2[(CH3)3CBr][N3-]}{k_{-1}[Br-] + k2[N3-]}}$


My thoughts: First of all, I read that $\ce{(CH3)3C+ (aq)}$ is a steady state concentration, which meant that $\frac{d\ce{[(CH3)3C+ (aq)}]}{dt} = 0,$ which prompted me to write out differential equations for each reactant. Thus, I got a system of differential equations:
$$\frac{d\ce{[(CH3)3CBr (aq)}]}{dt} =-\frac{k_1}{k_{-1}}\ce{[(CH3)3CBr (aq)}]$$
$$0=\frac{d\ce{[(CH3)3C+ (aq)}]}{dt} = -\frac{d\ce{[(CH3)3CBr (aq)}]}{dt} - \frac{d\ce{[(CH3)3CN3 (aq)}]}{dt} = \frac{k_1}{k_{-1}}\ce{[(CH3)3CBr (aq)}] -k_2 \frac{d\ce{[(CH3)3CN3 (aq)]}}{dt}$$
$$\frac{d\ce{[Br- (aq)}]}{dt} = -\frac{d\ce{[(CH3)3CBr (aq)}]}{dt} = \frac{k_1}{k_{-1}}\ce{[(CH3)3CBr (aq)}]$$
$$\frac{d\ce{[N3- (aq)}]}{dt} = -k_2\ce{[N3-(aq)]}$$

But I was unsure of how to use these differential equations to actually find the rate law. Could anyone provide any suggestions?
You really don't have to write all that differential equations. 

Note: I'll write all my steps on a paper, since MathJax formatting is a little time-taking at the moment.

Consider the rate laws for the two reactions:


Since the $\ce{[(CH3)3C+]}$ isn't changing with time, we get


From which, you get



Plugging these in the Rate2 which I mentioned, you will get,


Which is the same as what's given in option (D).

I hope I have satisfied your queries.