EXPLAIN THE KVL EQUATION IN T$>0.$ WHY IS IT $2(2$iA$)+4$iA$????$

Example $7\mathrm{.}5$: $v_o(t)$

$t<mn>0</mn><mi>L</mi>$ short under DC with switch position $1.$

$v_o(0_{-})=12V-(-2i_A)=12+2(\frac{12}{4})=18V$

$i_L(0_{-})=\frac{v_o(0_{-})}{6}=3A$

$t>0$ Turn independent sources OFF for Thiyenin resistance at $L($switch position $2)$:

$R_{TH}=2||4+\frac{2i_A}{3i_A}+6=\frac{4}{3}+\frac{2}{3}+6=8$

$=\frac{L}{R_{TH}}=3\frac{H}{8}=0\mathrm{.}375s$

BC with $L$ short:

$i_L()=\frac{4i_A-(-2i_A)}{6}=i_A$

$KVL$:$36V=2(2i_A)+4i_A=>i_A=4\mathrm{.}5A=i_L()$

Solutions with

matched $BCs$:

$v_o(0_{-})=6i_L(t)=\{\begin{array}{cc}18V,& t<mn>0</mn>\\ -9e^{-\frac{t}{}}+27V,& t>0\end{array}$