

In this modified astable multivibrator circuit the forward voltage drop of the diode D1 is always ignored, and the timing formulas are given as
$$ t_1 = \ln(2)R_1C_1, \quad t_2 = \ln(2)R_2C_1. $$
If you take \$R_1 = R_2\$ and ignore the diode voltage drop, you achieve %50 duty cycle. But, what happens if we don't ignore the diode voltage drop?
Suppose that the forward voltage drop of D1 is \$V_D\$. The voltage at pin 5 will be \$\dfrac{2}{3}V_{cc}\$.
The most general form of capacitor charging equation is
$$ v_c(t) = V_s + \left[ v_c(t_0) - V_s \right] e^{-\dfrac{t-t_0}{RC}}, \quad t\ge t_0. $$
If we rearrange the terms to get the time difference, we get
$$ \Delta t = t - t_0 = RC \ln \left[ \dfrac{V_s - v_c(t_0)}{V_s - v_c(t)} \right] . $$
Where, \$v_c(t)\$ is the function of capacitor voltage, \$V_s\$ is the source voltage.
During the on-time, the capacitor C1 will charge from \$\dfrac{1}{3}V_{cc}\$ to \$\dfrac{2}{3}V_{cc}\$ over R1. The supply voltage this R-C network sees is \$V_{cc}-V_D\$.
$$ t_1 = R_1C_1 \ln \left[ \dfrac{V_{cc} - V_D - \dfrac{1}{3}V_{cc}}{V_{cc} - V_D - \dfrac{2}{3}V_{cc}} \right] = R_1C_1 \ln \left[ \dfrac{\dfrac{2}{3}V_{cc} - V_D}{\dfrac{1}{3}V_{cc} - V_D} \right] = R_1C_1 \ln \left[ 1 + \dfrac{V_{cc}}{V_{cc} - 3V_D} \right] $$
During the discharge, D1 has no effect. So, similarly, \$t_2\$ will be
$$ t_2 = R_2C_1 \ln \left[ \dfrac{V_{cc} - \dfrac{1}{3}V_{cc}}{V_{cc} - \dfrac{2}{3}V_{cc}} \right] = \ln(2) R_2C_1. $$
Then the period of the oscillation is
$$ \boxed{T = t_1 + t_2 = R_1C_1 \ln \left[ 1 + \dfrac{V_{cc}}{V_{cc} - 3V_D} \right] + \ln(2) R_2C_1}. $$
And the frequency is
$$ \boxed{f = \dfrac{1}{T} = \dfrac{1}{t_1 + t_2} = \dfrac{1}{R_1C_1 \ln \left[ 1 + \dfrac{V_{cc}}{V_{cc} - 3V_D} \right] + \ln(2) R_2C_1}}. $$
(Adapted from this question.)
Compensation of The Error Due To Diode Forward Voltage Drop
We want to make the on and off times are equal. That is
$$ t_1 = t_2. $$
Then,
$$
R_1C_1 \ln \left[ 1 + \dfrac{V_{cc}}{V_{cc} - 3V_D} \right] = \ln(2) R_2C_1,\\
\ln \left[ 1 + \dfrac{V_{cc}}{V_{cc} - 3V_D} \right] = \dfrac{R_2}{R_1} \ln(2),\\
\boxed{R_1 = R_2 \dfrac{\ln(2)}{\ln \left[ 1 + \dfrac{V_{cc}}{V_{cc} - 3V_D} \right]}
\quad\quad \text{or} \quad\quad
R_2 = R_1 \dfrac{\ln \left[ 1 + \dfrac{V_{cc}}{V_{cc} - 3V_D} \right]}{\ln(2)}}.
$$
You have to choose \$R_1\$ and \$R_2\$ proportional to each other like this in order to achieve precisely symmetrical 50% duty cycle.