Finding stationary solutions and verifying differential equations Consider a trimolecular system where the following chemical reactions occur: 
  $$\ce{A + B -&gt;[k_1] A + C}\tag1$$
  $$\ce{A -&gt;[k_2] B}\tag2$$
  $$\ce{C -&gt;[k_3] C}\tag3$$
  Write the differential equations that model the evolution of the three component molar concentrations and determine the stationary solutions (there are two). Finally, if initially $[\ce{A}]=0.75$, $[\ce{B}]=0.25$ and $[\ce{C}]=0$, calculate the time when the mole fraction of $\ce{A}$ is half of its initial value. 


I think that the differential equations, following the expression for the reaction velocity, are
$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t}=-k_2[\ce{A}]$$ 
$$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}=-k_1[\ce{A}][\ce{B}]+k_2[\ce{A}]$$
$$\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t}=k_1[\ce{A}][\ce{B}]$$
However, I don't know how to proceed with the stationary solutions. I guess that one of them would be obtained doing:
$$-k_1[\ce{A}][\ce{B}]+k_2[\ce{A}]=0$$
but I'm not sure.

Finally, I got the relations 
$$[\ce{B}]=\left(\frac{1}{4}-\frac{k_2}{k_1}\right)\exp\left(\tfrac{k_1}{k_2}(-\tfrac{3}{4}+[\ce{A}])\right)+\frac{k_2}{k_1}$$
$$[\ce{C}]=-\left(\frac{1}{4}-\frac{k_2}{k_1}\right)\exp\left(\tfrac{k_1}{k_2}(-\tfrac{3}{4}+[\ce{A}])\right)+1-\frac{k_2}{k_1}-[\ce{A}]$$
but I have no idea how to obtain the time. Any help would be appreciated.
Your differential equations are correct.

From $\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = -k_1[\ce{A}][\ce{B}]+k_2[\ce{A}] = 0$ follow the stationary solutions $[\ce{A}] = 0$ or $[\ce{B}] = \frac{k_2}{k_1}$.

Reaction (2) obeys a first order decay law. Its half-life is $t_{\frac{1}{2}} = \mathrm{ln}(2) / k_2$.

The image below shows how $[\ce{B}]$ reaches the stationary state soon after the reaction has started ($k_1 = 0.3, k_2 = 0.2$).