System of Differential Equations- Asymmetric First-Price Auction
I followed Oliv's suggestion, which was quite fruitful. So we have the differential equations:

$$g_{1}^{\prime}(b) = \dfrac{1}{2} \cdot \dfrac{g_{1}(b) - 1}{g_{2}(b) - b}$$

And:

$$g_{2}^{\prime}(b) = \dfrac{3}{2} \cdot \dfrac{1}{g_{1}(b) - b}$$

With the boundary conditions $g_{2}(0) = 0$ and $g_{1}(\overline{b}) = g_{2}(\overline{b}) = 3$, where $\overline{b}$ is the maximum bid. 

Now we guess that $g_{1}(b) = \alpha b + \gamma$ and $g_{2}(b) = \delta b + \lambda$. Applying $g_{2}(b) = 0$ yields that $\lambda = 0$. 

Next, I substitute $g_{1}(b)$ into $g_{2}^{\prime}(b)$ to obtain:

$$g_{2}^{\prime}(b) = \dfrac{3}{2} \cdot \dfrac{1}{(\alpha - 1)b + \gamma}$$

Integrating $g_{2}^{\prime}(b)$ yields 

$$g_{2}(b) = \dfrac{3}{2(\alpha - 1)} ln( (\alpha - 1)b + \gamma)$$

We note there is no constant when integrating, as $g_{2}(b) = \delta b$. We now apply $g_{2}(0) = 0$ again, concluding that $ln( \gamma) = 0$. And so $\gamma = 1$. Thus, $g_{1}(b) = \alpha b + 1$.

We now solve:

$$g_{2}(\overline{b}) = 3 = \dfrac{3}{2(\alpha - 1)} ln( (\alpha - 1)\overline{b} + \gamma)$$

From this and noting that $g_{1}(\overline{b}) = 3 = \alpha \overline{b} + 1$, we obtain:

$$e^{2(\alpha - 1)} = 3 - \overline{b} \implies \overline{b} = 3 - e^{2(\alpha - 1)}$$

Plugging this into $g_{1}(b)$ yields:

$$g_{1}(\overline{b}) = \alpha(3 - e^{2(\alpha - 1)}) + 1 = 3$$ 

Which implies that the solution $\alpha = 1$. Thus, $\overline{b} = 2$.

So $\delta = \dfrac{3}{2}$. 

Thus, $g_{1}(b) = b + 1 \implies \beta_{1}(v) = v - 1$; and $g_{2}(b) = \dfrac{3}{2}b$ implies $\beta_{2}(v) = \dfrac{2}{3}v$ as desired.