91Ó°ÊÓ

One will find a particular solution using the method of undetermined coefficients.

Assume that a particular solution for \(q\) has a form of\({q_p}(t) = A{e^{i\gamma t}}\)

Then

Substituting this into the given differential equation one will obtain a value for the constant\(A\).

\(\begin{aligned}{c}L\frac{{{d^2}{q_p}}}{{d{t^2}}}{\bf{ + }}R\frac{{d{q_p}}}{{dt}}{\bf{ + }}\frac{1}{C}{q_p}{\bf{ = }}L\left( {{\bf{ - }}{\gamma ^2}A{e^{i\gamma t}}} \right){\bf{ + }}R \cdot i\gamma A{e^{i\gamma t}}{\bf{ + }}\frac{1}{C}A{e^{i\gamma t}}\\{\bf{ = }}A\left( {{\bf{ - }}{\gamma ^2}L{\bf{ + }}iR\gamma {\bf{ + }}\frac{1}{C}} \right){e^{i\gamma t}}\\{\bf{ = }}{E_0}{e^{i\gamma t}}\end{aligned}\)

Dividing the previous equation by \({e^{i\gamma t}}\)one will get

\(\begin{aligned}{c}A\left( {{\bf{ - }}{\gamma ^2}L + iR\gamma {\bf{ + }}\frac{1}{C}} \right){\bf{ = }}{E_0}\\A{\bf{ = }}\frac{{{E_0}}}{{\frac{1}{C}{\bf{ - }}{\gamma ^2}L{\bf{ + }}i\gamma R}}\end{aligned}\)

Substituting the derived value for \(A\)one has that the particular solution for \(q\) is\({q_p}(t){\bf{ = }}\frac{{{E_0}}}{{\frac{1}{C}{\bf{ - }}{\gamma ^2}L{\bf{ + }}i\gamma R}}{e^{i\gamma t}}\).

The corresponding differential equation for the current is:

\(\begin{aligned}{c}L\frac{{{d^2}I}}{{d{t^2}}}{\bf{ + }}R\frac{{dI}}{{dt}}{\bf{ + }}\frac{1}{C}I{\bf{ = }}\frac{d}{{dt}}\left( {{E_0}{e^{i\gamma t}}} \right)\\L\frac{{{d^2}I}}{{d{t^2}}} + R\frac{{dI}}{{dt}}{\bf{ + }}\frac{1}{C}I{\bf{ = }}i\gamma {E_0}{e^{i\gamma t}}\end{aligned}\)

One will use the method of undetermined coefficients to find the steady-state current.

Assume that\({I_p}(t) = B{e^{i\gamma t}}\). Then

\(I_p {}'(r) = i\gamma B{e^{i\gamma t}},\;\;\;I_p {}''(t){\bf{ = - }}{\gamma ^2}B{e^{i\gamma t}}\).

As before, one will substitute this into the differential equation for current and obtain a value for\(B\):

\(\begin{aligned}{c}L\frac{{{d^2}{I_p}}}{{d{t^2}}}{\bf{ + }}R\frac{{d{I_p}}}{{dt}}{\bf{ + }}\frac{1}{C}{I_p}{\bf{ = }}L\left( {{\bf{ - }}{\gamma ^2}B{e^{i\gamma t}}} \right){\bf{ + }}R \cdot i\gamma B{e^{i\gamma t}}{\bf{ + }}\frac{1}{C}B{e^{i\gamma t}}\\{\bf{ = }}B\left( {{\bf{ - }}{\gamma ^2}L{\bf{ + }}iR\gamma {\bf{ + }}\frac{1}{C}} \right){e^{i\gamma t}}\\{\bf{ = }}i\gamma {E_0}{e^{i\gamma t}}\end{aligned}\)

Dividing the previous equation by \({e^{i\gamma t}}\)one will get a value for \(B\):

\(\begin{aligned}{c}B\left( { - {\gamma ^2}L{\bf{ + }}iR\gamma {\bf{ + }}\frac{1}{C}} \right){\bf{ = }}i\gamma {E_0}\\B{\bf{ = }}\frac{{i\gamma {E_0}}}{{\frac{1}{C} - {\gamma ^2}L + i\gamma R}}\\B{\bf{ = }}\frac{{{E_0}}}{{\frac{1}{{i\gamma C}}{\bf{ - }}\frac{{{\gamma ^2}L}}{{i\gamma }}{\bf{ + }}\frac{{i\gamma R}}{{i\gamma }}}}\\B = \frac{{{E_0}}}{{R{\bf{ + }}i(\gamma L{\bf{ - }}\frac{1}{{(\gamma C)}})}}\end{aligned}\)

And now substituting the value for \(B\) we will get that the steady-state current is\({I_p}(t){\bf{ = }}\frac{{{E_0}}}{{R{\bf{ + }}i(\gamma L{\bf{ - }}\frac{1}{{(\gamma C)}})}}{e^{i\gamma t}}\)

One will transform the steady-state current that one obtains in part\(\left( {\bf{b}} \right)\):

\(\begin{aligned}{c}{I_p}(t){\bf{ = }}\frac{{{E_0}}}{{R{\bf{ + }}i(\gamma L{\bf{ - }}\frac{1}{{(\gamma C)}})}}{e^{i\gamma t}}\\{\bf{ = }}\frac{{{E_0}(R{\bf{ - }}i(\gamma L{\bf{ - }}\frac{1}{{(\gamma C)}})}}{{(R{\bf{ + }}i(\gamma L{\bf{ - }}\frac{1}{{(\gamma C)}}))(R{\bf{ - }}i(\gamma L{\bf{ - }}\frac{1}{{(\gamma C)}}))}}{e^{i\gamma t}}\\{\bf{ = }}\frac{{{E_0}(R{\bf{ - }}i(\gamma L{\bf{ - }}\frac{1}{{(\gamma C)}})}}{{{R^2}{\bf{ + }}{{(\gamma L{\bf{ - }}\frac{1}{{(\gamma C)}})}^2}}}{e^{i\gamma t}}\end{aligned}\)

One can apply the given relation\(\alpha {\bf{ + }}i\beta {\bf{ = }}\sqrt {{\alpha ^2}{\bf{ + }}{\beta ^2}} {e^{i\theta }}\), where \(\tan \theta {\bf{ = }}\frac{\beta }{\alpha }\) for \(\alpha {\bf{ = }}R\) and\(\beta {\bf{ = - }}(\gamma L{\bf{ - }}\frac{1}{{(\gamma C)}}){\bf{ = }}\frac{1}{{(\gamma C)}}{\bf{ - }}\gamma L\). So, one has

\(\begin{aligned}{c}{I_p}(t){\bf{ = }}\frac{{{E_0}\sqrt {{R^2}{\bf{ + }}{{(\frac{1}{{(\gamma C)}} - \gamma L)}^2}} {e^{i\theta }}}}{{{R^2}{\bf{ + }}{{(\gamma L{\bf{ - }}\frac{1}{{(\gamma C)}})}^2}}}{e^{i\gamma t}}\\{\bf{ = }}\frac{{{E_0}}}{{\sqrt {{R^2}{\bf{ + }}{{(\gamma L{\bf{ - }}\frac{1}{{(\gamma C)}})}^2}} }}{e^{i(\gamma t{\bf{ + }}\theta )}}\end{aligned}\),

Where\(\tan \theta {\bf{ = }}\frac{{\frac{1}{{(\gamma C)}} - \gamma L}}{R}{\bf{ = }}\frac{{\frac{1}{C}{\bf{ - }}L{\gamma ^2}}}{{\gamma R}}\).

Let's rewrite the steady-state current obtained in part \(\left( {\bf{c}} \right)\)

\(\begin{aligned}{c}{I_p}(t){\bf{ = }}\frac{{{E_0}}}{{\sqrt {{R^2}{\bf{ + }}{{(\gamma L{\bf{ - }}\frac{1}{{(\gamma C)}})}^2}} }}{e^{i(\gamma t{\bf{ + }}\theta )}}\\{\bf{ = }}\frac{{{E_0}}}{{\sqrt {{R^2}{\bf{ + }}{{(\gamma L{\bf{ - }}\frac{1}{{(\gamma C)}})}^2}} }}(\cos (\gamma t{\bf{ + }}\theta ) + i\sin (\gamma t{\bf{ + }}\theta ))\end{aligned}\)

Where\(\tan \theta = \frac{{\frac{1}{C}{\bf{ - }}L{\gamma ^2}}}{{\gamma R}}\). Now it is easy to see that the imaginary part of \({I_p}\) is\({\mathop{\rm Im}\nolimits} \left( {{I_p}(t)} \right){\bf{ = }}\frac{{{E_0}\sin (\gamma t + \theta )}}{{\sqrt {{R^2}{\bf{ + }}{{(\gamma L{\bf{ - }}\frac{1}{{(\gamma C)}})}^2}} }}\)where \(\tan \theta {\bf{ = }}\frac{{\frac{1}{C}{\bf{ - }}L{\gamma ^2}}}{{\gamma R}}\)which is identical to the solution given in the equation \(\left( {{\bf{10}}} \right)\)in the textbook, that one wants to show.