Let the competitive reactions be $A \stackrel{k_1}{\rightarrow}B$ and $A \stackrel{k_2}{\rightarrow}C$ of first order, we will determine $[A]$, $[B]$ and $[ C]$ as a function of time, assuming $[B]_0=[C]_0=0$. The variation of $[A]$ in time is given by the expression:

$$\frac{d[A]}{dt}=-k_1[A]-k_2[A]\;\;\rightarrow \frac{d[A]}{dt}=(-k_1-k_2)[A]$$

Separating variables and integrating:

$$[A]=[A]_0 e^{-( k_1+k_2)t} \label{ec1}$$

We now proceed to obtain the variation of $[B]$ with t

$$\frac{d[B]}{dt}=k_1[A]$$

Substituting $[A]$ for $(\ref{ec1})$

$$\frac{d[B]}{dt}=k_1[A]_0 e^{-(k_1+k_2)t}$$

Separating variables and integrating

$$[B]=\left[\frac{k_1[A]_0}{-(k_1+k_2)}^{-(k_1+k_2)t}\right ]_{0}^{t}$$

Substituting the limits of integration

$$[B]=\frac{k_1 [A]_0}{k_1 +k_2}\left[1-e^{ -(k_1+k_2)t}\right]$$

The variation of $[C]$ with t follows analogously from the differential equation $\frac{d[C]}{dt}= k_2[A]$

$$[C]=\frac{k_2[A]_0}{k_1 +k_2}\left[1-e^{-(k_1+k_2)t}\right]$$

Dividing both concentrations at any inst before time we get:

$$\frac{[C]}{[B]}=\frac{k_2}{k_1}$$