An \({\bf{R L C}}\) series circuit has a voltage source given by \({\bf{E(t) = 40cos2t\;V}}\), a resistor of \({\bf{2\Omega }}\), an inductor of \({\bf{1/4H}}\), and a capacitor of \({\bf{1/13\;F}}\). If the initial current is zero and the initial charge on the capacitor is \({\bf{3}}{\bf{.5C}}\), determine the charge on the capacitor for \({\bf{t > 0}}\).

The differential equation we need to solve is

\(\begin{aligned}{c}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\bf{L}}\frac{{{\bf{dI}}}}{{{\bf{dt}}}}{\bf{ + RI + }}\frac{{\bf{1}}}{{\bf{C}}}{\bf{q = E}}\\{\bf{L}}\frac{{{{\bf{d}}^{\bf{2}}}{\bf{q}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ + R}}\frac{{{\bf{dq}}}}{{{\bf{dt}}}}{\bf{ + }}\frac{{\bf{1}}}{{\bf{C}}}{\bf{q = E}}\end{aligned}\)

One has that\({\bf{E(t) = 40cos2t\;V,R = 2\Omega ,L = }}\frac{{\bf{1}}}{{\bf{4}}}{\bf{H,C = }}\frac{{\bf{1}}}{{{\bf{13}}}}{\bf{\;F}}\), and substituting this into the previous differential equation one will get

\(\begin{aligned}{c}\frac{{\bf{1}}}{{\bf{4}}}{\bf{q}}''{\bf{ + 2q}}'{\bf{ + 13q = 40cos2t}}\\{\bf{q}}''{\bf{ + 8q}}'{\bf{ + 52q = 160cos2t}}\end{aligned}\)

First, one will find a homogeneous solution for\({\bf{q}}\left( {\bf{t}} \right)\). The auxiliary equation is \({{\bf{r}}^{\bf{2}}}{\bf{ + 8r + 52 = 0}}\) and its solutions are

\(\begin{aligned}{c}{{\bf{r}}_{{\bf{1,2}}}}{\bf{ = }}\frac{{{\bf{ - 8 \pm }}\sqrt {{\bf{64 - 208}}} }}{{\bf{2}}}\\{\bf{ = }}\frac{{{\bf{ - 8 \pm }}\sqrt {{\bf{ - 144}}} }}{{\bf{2}}}\\{\bf{ = }}\frac{{{\bf{ - 8 \pm 12i}}}}{{\bf{2}}}\\{{\bf{r}}_{{\bf{1,2}}}}{\bf{ = - 4 \pm 6i}}\end{aligned}\)

So, the homogeneous solution is\({{\bf{q}}_{\bf{h}}}{\bf{(t) = }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - 4t}}}}{\bf{cos6t + }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - 4t}}}}{\bf{sin6t}}\).

One will find a particular solution by\({\bf{q}}\) using the method of undetermined coefficients. Assume that\({{\bf{q}}_{\bf{p}}}{\bf{(t) = Acos2t + Bsin2t}}\).

Now,one has that

\(\begin{aligned}{c}{\bf{q}}_{\bf{p}}^{}{\bf{'(t) = - 2Asin2t + 2Bcos2t,}}\\\;{\bf{q}}_{\bf{p}}^{}{\bf{''(t) = - 4Acos2t - 4Bsin2t}}\end{aligned}\)

Substituting this into the differential equation one has that

\(\begin{aligned}{c}{\bf{q}}_{\bf{p}}^{}{\bf{''(t) + 8q}}_{\bf{p}}^{}{\bf{'(t) + 52}}{{\bf{q}}_{\bf{p}}}{\bf{(t) = - 4Acos2t - 4Bsin2t + 8( - 2Asin2t + 2Bcos2t) + 52(Acos2t + Bsin2t)}}\\{\bf{ = (48A + 16B)cos2t + (48B - 16A)sin2t}}\\{\bf{ = 16(3A + B)cos2t - 16(A - 3B)sin2t}}\\{\bf{ = 160cos2t}}\end{aligned}\)

Equating the multiplying coefficients, one has that\({\bf{16}}\left( {{\bf{3 A + B}}} \right){\bf{ = 160}}\)and \({\bf{ - 16}}\left( {{\bf{A - 3 B}}} \right){\bf{ = 0}}\)

\({\bf{3 A + B = 10}}\)and\({\bf{A - 3 B = 0}}\).

The second equation gives us that \({\bf{A = 3 B}}\) and substituting this into the first equation one has that \({\bf{9B + B = 10}} \Rightarrow {\bf{B = 1}}\) and therefore\({\bf{A = 3B = 3}}\).

So, the particular solution for \({\bf{q}}\left( {\bf{t}} \right)\) is\({{\bf{q}}_{\bf{p}}}{\bf{(t) = 3cos2t + sin2t}}\)and the general solution is\(\begin{aligned}{c}{\bf{q(t) = }}{{\bf{q}}_{\bf{h}}}{\bf{(t) + }}{{\bf{q}}_{\bf{p}}}{\bf{(t)}}\\{\bf{ = }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - 4t}}}}{\bf{cos6t + }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - 4t}}}}{\bf{sin6t + 3cos2t + sin2t}}\end{aligned}\)

The initial conditions are \({\bf{I(0) = q'(0) = 0}}\) and \({\bf{q(0) = 3}}{\bf{.5C}}\). The general solution for current is\({\bf{I(t) = q}}'{\bf{(t) = - 4}}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - 4t}}}}{\bf{cos6t - 6}}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - 4t}}}}{\bf{sin6t - 4}}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - 4t}}}}{\bf{sin6t + 6}}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - 4t}}}}{\bf{cos6t - 6sin2t + 2cos2t}}\).

From the second initial condition, one has that\({\bf{q(0) = }}{{\bf{c}}_{\bf{1}}}{\bf{ + 3 = 3}}{\bf{.5}}\; \Leftrightarrow {{\bf{c}}_{\bf{1}}}{\bf{ = 0}}{\bf{.5}}\)

Substituting the value for \({{\bf{c}}_{\bf{1}}}\) into the general solution for the current one will get that;

\(\begin{aligned}{c}{\bf{I(t) = - 2}}{{\bf{e}}^{{\bf{ - 4t}}}}{\bf{cos6t - 3}}{{\bf{e}}^{{\bf{ - 4t}}}}{\bf{sin6t - 4}}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - 4t}}}}{\bf{sin6t + 6}}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - 4t}}}}{\bf{cos6t - 6sin2t + 2cos2t}}\\{\bf{I(0) = - 2 + 6}}{{\bf{c}}_{\bf{2}}}{\bf{ + 2 = 0}} \Leftrightarrow {{\bf{c}}_{\bf{2}}}{\bf{ = 0}}\end{aligned}\)

Finally, the charge on the capacitor is given by\({\bf{q(t) = 0}}{\bf{.5}}{{\bf{e}}^{{\bf{ - 4t}}}}{\bf{cos6t + 3cos2t + sin2t}}\)