I get\[L^{2}Ci'''_R + RLCi''_R + Li'_R = e(t)\]for the time-domain equation.
Since we are taking Laplace transforms of derivatives up to third-order,
\[i''_R(0), i'_R(0),\]and\[i_R(0)\]enter into my calculations. Since nothing
is stated about initial values for the current across the resistor, I keep those
three numbers arbitrary, leaving them in symbolic form. When I take the
Laplace transform of both sides of the equation I came up with,
I get
\[I_R(s) = E(s)/(L^{2}Cs^{3}+ RLCs^{2} + Ls)\]
\[+ L^{2}Ci_R(0)s^{2}/(L^{2}Cs^{3} + RLCs^{2}+ Ls)\]
\[+ (L^{2}Ci'_R(0) + RLCi_R(0))s/(L^{2}Cs^{3} + RLCs^{2} + Ls)\]
\[+ (L^{2}Ci''_R(0) + RLCi'_R(0) + LI_R(0))/(L^{2}Cs^{3} + RLCs^{2} + Ls).\]I get\[L^{2}Ci'''_R + RLCi''_R + Li'_R = e(t)\]for the time-domain equation.
Since we are taking Laplace transforms of derivatives up to third-order,
\[i''_R(0), i'_R(0),\]and\[i_R(0)\]enter into my calculations. Since nothing
is stated about initial values for the current across the resistor, I keep those
three numbers arbitrary, leaving them in symbolic form. When I take the
Laplace transform of both sides of the equation I came up with,
I get
\[I_R(s) = E(s)/(L^{2}Cs^{3}+ RLCs^{2} + Ls)\]
\[+ L^{2}Ci_R(0)s^{2}/(L^{2}Cs^{3} + RLCs^{2}+ Ls)\]
\[+ (L^{2}Ci'_R(0) + RLCi_R(0))s/(L^{2}Cs^{3} + RLCs^{2} + Ls)\]
\[+ (L^{2}Ci''_R(0) + RLCi'_R(0) + LI_R(0))/(L^{2}Cs^{3} + RLCs^{2} + Ls).\]