How can I solve for the DC analysis of the CE transistor?
This type of calculation is not easy because there are too many unknowns in the circuit.
For example, we do not know the exact value of a \$V_{BE}\$ voltage, and also the \$\beta\$ value is also unknown.
But if we assumed the \$\beta = \infty\$ and some constant \$V_{BE}\$ value
We can write this KVL for this simplified schematic:



simulate this circuit &ndash; Schematic created using CircuitLab
$$V_{CC} = I_{C1}R_C + V_{BE2} + I_DR_1+I_DR_2$$
$$I_DR_2=V_{BE1}+I_{C1} R_{E1} $$
And if we solve it we get this:
$$I_{C1} = \frac{ V_{CC} - \left(\frac{R_1 + 2R_2}{R_2}V_{BE}\right)}{ R_{E1} + R_{C1} +\frac{R_1}{R_2}R_{E1}} $$
And knowing that \$V_E = V_C - V_{BE2} = V_{CC} - I_{C1}R_{C1} - V_{BE2}\$
And \$V_B = V_E \times \frac{R_2}{R_1+R_2}\$
We have:
$$V_E = \frac{\left( \frac{R_{E1}}{R_{C1}} \times(V_{CC} - V_{BE})\right)+V_{BE}}{\frac{R_2}{R_1 + R_2} + \frac{R_{E1}}{R_{C1}}}$$
$$ V_B = \frac{V_{CC} + V_{BE}\left(\frac{R_{C1}}{R_{E1}} -1\right)}{\left(1+\frac{R_1}{R_2}\right) +\frac{R_{C1}}{R_{E1}}}$$
Doing such a calculation does not have much sense. But here you have it.