Since \(I_G \simeq 0\ A\) and \(I_D = I_S\)
\(\boxed{I_{R_1} = I_{R_2} = \frac{V_{DD}}{R_1 + R_2}}\)
\(\boxed{V_G = \frac{V_{DD} R_2}{R_1 + R_2}}\)
\(\displaystyle -V_G + V_{GS} + I_S R_S = 0\) \(\displaystyle -V_G + V_{GS} + I_D R_S = 0\)
\(\boxed{V_{GS} = V_G - I_D R_S}\)
For \(V_{GS} > V_{GS(Th)}\), substitute \(V_{GS}\) to the transfer curve equation:
\(\boxed{k = \frac{I_{D(on)}}{\left( V_{GS(on)} - V_{GS(Th)} \right) ^ 2}}\)
\(I_D = k \left( V_{GS} - V_{GS(Th)} \right) ^ 2\)
\(\boxed{I_D = k \left( V_G - I_D R_S - V_{GS(Th)} \right) ^ 2}\)
\(-V_{DD} + I_D R_D + V_{DS} + I_S R_S = 0\) \(-V_{DD} + I_D R_D + V_{DS} + I_D R_S = 0\)
\(\boxed{V_{DS} = V_{DD} - I_D \left( R_D + R_S \right)}\)
\(\boxed{V_D = V_{DD} - I_D R_D}\)
\(\boxed{V_S = I_D R_S}\)