Fig.2. RLC-circuit.
Problem 2.
Determine $I(t)$ for (a) a constant electromotive force,
(b) a periodic electromotive force,
if $I(0):=I_0$ and $I'(0)=I_0'$ are given.
we get
\[
LI'(t)+RI(t)+\frac{1}{C}\int I(t)\,dt=E(t).
\]
Differentiating the equation with respect to $t$ we find
\[
LI''(t)+RI'(t)+\frac{1}{C}I(t)=E'(t).
\]
Case (a). Constant electromotive force.
If $E(t)$ is constant, then it follows \[ LI''(t)+RI'(t)+\frac{1}{C}I(t)=0. \] The general solution of the homogeneous equation is (we make a detailed discussion below) \[ I(t)=c_1 e^{\lambda_1 t}+c_2 e^{\lambda_2 t}, \] where $\lambda_1$ and $\lambda_2$ are the roots of the characteristic equation \[ \lambda^2+\frac{R}{L}\lambda+\frac{1}{LC}=0, \] that is, \[ \lambda_{1,2}=-\frac{R}{2L}\pm\frac{\sqrt{R^2 C -4L}}{2L\sqrt{C}}. \] (i) $\boldsymbol{R^2 C -4L\gt 0}$.
Since $I(0)=I_0$, $I'(0)=I_0'$ we obtain the linear system \begin{align*} c_1+c_2 &=I_0,\\ \lambda_1 c_1+\lambda_2 c_2 &=I_0'. \end{align*} Solving this system \[ c_1=\frac{-I_0\lambda_2+I_0'}{\lambda_1-\lambda_2},\qquad c_2=-\frac{-I_0\lambda_1+I_0'}{\lambda_1-\lambda_2}. \] (ii) $\boldsymbol{R^2 C -4L= 0}$.
In this case \[ \lambda_1=\lambda_2=-\frac{R}{2L}, \] and \[ I(t)=(c_1+c_2 t)e^{-\frac{R}{2L}t}. \] Since $I(0)=I_0$, $I'(0)=I_0'$ we obtain the linear system \begin{align*} c_1 &=I_0,\\ c_2-\frac{c_1 R}{2L} &=I_0'. \end{align*} Solving this system \[ c_1=I_0,\qquad c_2=\frac{I_0 R}{2L}+I_0'. \] Hence \[ I(t)=\left(I_0+\left(\frac{I_0 R}{2L}+I_0' \right)t \right)e^{-\frac{R}{2L}t}. \] (iii) $\boldsymbol{R^2 C -4L\lt 0}$.
In this case \[ \lambda_{1,2}=-\frac{R}{2L}\pm i\frac{\sqrt{ 4L-R^2 C}}{2L\sqrt{C}}. \] The solution of the differential equation can be written in the form \[ I(t)=e^{-\frac{R}{2L}t}\left( d_1\sin\left(\frac{\sqrt{ 4L-R^2 C}}{2L\sqrt{C}}t \right)+d_2\cos\left(\frac{\sqrt{ 4L-R^2 C}}{2L\sqrt{C}}t \right) \right). \] Since $I(0)=I_0$, $I'(0)=I_0'$ we obtain the linear system \begin{align*} d_2 &=I_0,\\ -\frac{R d_2}{2L}+\frac{d_1\sqrt{-CR^2+4L}}{L\sqrt{C}} &=I_0'. \end{align*} Solving this system \[ d_2=I_0,\qquad d_1=\frac{\sqrt{C}(2LI_0'+RI_0)}{2\sqrt{-CR^2+4L}}. \] Hence \[ I(t)\!=\!e^{-\frac{R}{2L}t}\!\left(\!\! \frac{\sqrt{C}(2LI_0'\!+\!RI_0)}{2\sqrt{-CR^2\!+\!4L}}\sin\left(\!\frac{\sqrt{ 4L\!-\!R^2 C}}{2L\sqrt{C}}t\! \right)\!+\!I_0\cos\left(\!\frac{\sqrt{ 4L\!-\!R^2 C}}{2L\sqrt{C}}t\! \right)\!\! \right). \] Here we can transform the term in the bracket further if we use that \[ \frac{a\sin(x)+b\cos(x)}{\sqrt{a^2+b^2}}=\sin(x+\delta),\qquad(a,b\gt 0), \] where $\tan(\delta)=\frac{b}{a} $.
Hence \[ I(t)=\frac{e^{-\frac{R}{2L}t}}{\sqrt{\frac{C(2LI_0'+RI_0)^2}{4(-CR^2+4L)}+I_0^2}}\sin\left(\frac{\sqrt{ 4L-R^2 C}}{2L\sqrt{C}}t+\delta \right), \] where $\tan(\delta)=\frac{2I_0\sqrt{-CR^2+4L}}{\sqrt{C}(2LI_0'+RI_0)}$.
Case (b). Periodic electromotive force $E(t)=E_0\sin(\omega t)$.
In this case $E'(t)=E_0\omega\cos(\omega t)$ and \begin{equation} \label{eq:periodicdiffeq} LI''(t)+RI'(t)+\frac{1}{C}I(t)=E_0\omega\cos(\omega t). \end{equation} We seek a particular solution by the method of undetermined coefficients. Let \[ I_p(t):=a\cos(\omega t)+b\sin(\omega t). \] Then \begin{align*} I_p'(t) &=\omega(-a\sin(\omega t)+b\cos(\omega t)),\\ I_p''(t) &=\omega^2(-a\cos(\omega t)-b\sin(\omega t)). \end{align*} Substitute them into \eqref{eq:periodicdiffeq} we have
Then we collect the sine and cosine terms and equate them to $0$ and $E_0\omega\cos(\omega t)$, \begin{align*} L\omega^2(-a)+R\omega b+a/C &=E_0\omega,\\ L\omega^2(-b)+R\omega(-a)+b/C &=0. \end{align*} The solution of this system is \[ a=\frac{-E_0S}{R^2+S^2}, \qquad b=\frac{E_0R}{R^2+S^2}, \] where \[ S=\omega L-\frac{1}{\omega C}. \] Here we may assume that $R\gt 0$, so $R^2+S^2\gt 0$.
Hence \[ I_p(t)=\frac{E_0}{\sqrt{R^2+S^2}}\left(\frac{R}{\sqrt{R^2+S^2}}\sin(\omega t)-\frac{S}{\sqrt{R^2+S^2}}\cos(\omega t)\right). \] Here we can transform the first term further if we use that \[ \frac{A\sin(x)-B\cos(x)}{\sqrt{A^2+B^2}}=\sin(x-\delta),\qquad(A,B\gt 0), \] where $\tan(\delta)=\frac{B}{A} $.
Applying to our case \[ I_p(t)=I_0\sin(\omega t-\delta), \] where \[ I_0=\frac{E_0}{\sqrt{R^2+S^2}}, \] and \[ \tan(\delta)=\frac{S}{R}. \]
If $E(t)$ is constant, then it follows \[ LI''(t)+RI'(t)+\frac{1}{C}I(t)=0. \] The general solution of the homogeneous equation is (we make a detailed discussion below) \[ I(t)=c_1 e^{\lambda_1 t}+c_2 e^{\lambda_2 t}, \] where $\lambda_1$ and $\lambda_2$ are the roots of the characteristic equation \[ \lambda^2+\frac{R}{L}\lambda+\frac{1}{LC}=0, \] that is, \[ \lambda_{1,2}=-\frac{R}{2L}\pm\frac{\sqrt{R^2 C -4L}}{2L\sqrt{C}}. \] (i) $\boldsymbol{R^2 C -4L\gt 0}$.
Since $I(0)=I_0$, $I'(0)=I_0'$ we obtain the linear system \begin{align*} c_1+c_2 &=I_0,\\ \lambda_1 c_1+\lambda_2 c_2 &=I_0'. \end{align*} Solving this system \[ c_1=\frac{-I_0\lambda_2+I_0'}{\lambda_1-\lambda_2},\qquad c_2=-\frac{-I_0\lambda_1+I_0'}{\lambda_1-\lambda_2}. \] (ii) $\boldsymbol{R^2 C -4L= 0}$.
In this case \[ \lambda_1=\lambda_2=-\frac{R}{2L}, \] and \[ I(t)=(c_1+c_2 t)e^{-\frac{R}{2L}t}. \] Since $I(0)=I_0$, $I'(0)=I_0'$ we obtain the linear system \begin{align*} c_1 &=I_0,\\ c_2-\frac{c_1 R}{2L} &=I_0'. \end{align*} Solving this system \[ c_1=I_0,\qquad c_2=\frac{I_0 R}{2L}+I_0'. \] Hence \[ I(t)=\left(I_0+\left(\frac{I_0 R}{2L}+I_0' \right)t \right)e^{-\frac{R}{2L}t}. \] (iii) $\boldsymbol{R^2 C -4L\lt 0}$.
In this case \[ \lambda_{1,2}=-\frac{R}{2L}\pm i\frac{\sqrt{ 4L-R^2 C}}{2L\sqrt{C}}. \] The solution of the differential equation can be written in the form \[ I(t)=e^{-\frac{R}{2L}t}\left( d_1\sin\left(\frac{\sqrt{ 4L-R^2 C}}{2L\sqrt{C}}t \right)+d_2\cos\left(\frac{\sqrt{ 4L-R^2 C}}{2L\sqrt{C}}t \right) \right). \] Since $I(0)=I_0$, $I'(0)=I_0'$ we obtain the linear system \begin{align*} d_2 &=I_0,\\ -\frac{R d_2}{2L}+\frac{d_1\sqrt{-CR^2+4L}}{L\sqrt{C}} &=I_0'. \end{align*} Solving this system \[ d_2=I_0,\qquad d_1=\frac{\sqrt{C}(2LI_0'+RI_0)}{2\sqrt{-CR^2+4L}}. \] Hence \[ I(t)\!=\!e^{-\frac{R}{2L}t}\!\left(\!\! \frac{\sqrt{C}(2LI_0'\!+\!RI_0)}{2\sqrt{-CR^2\!+\!4L}}\sin\left(\!\frac{\sqrt{ 4L\!-\!R^2 C}}{2L\sqrt{C}}t\! \right)\!+\!I_0\cos\left(\!\frac{\sqrt{ 4L\!-\!R^2 C}}{2L\sqrt{C}}t\! \right)\!\! \right). \] Here we can transform the term in the bracket further if we use that \[ \frac{a\sin(x)+b\cos(x)}{\sqrt{a^2+b^2}}=\sin(x+\delta),\qquad(a,b\gt 0), \] where $\tan(\delta)=\frac{b}{a} $.
Hence \[ I(t)=\frac{e^{-\frac{R}{2L}t}}{\sqrt{\frac{C(2LI_0'+RI_0)^2}{4(-CR^2+4L)}+I_0^2}}\sin\left(\frac{\sqrt{ 4L-R^2 C}}{2L\sqrt{C}}t+\delta \right), \] where $\tan(\delta)=\frac{2I_0\sqrt{-CR^2+4L}}{\sqrt{C}(2LI_0'+RI_0)}$.
Case (b). Periodic electromotive force $E(t)=E_0\sin(\omega t)$.
In this case $E'(t)=E_0\omega\cos(\omega t)$ and \begin{equation} \label{eq:periodicdiffeq} LI''(t)+RI'(t)+\frac{1}{C}I(t)=E_0\omega\cos(\omega t). \end{equation} We seek a particular solution by the method of undetermined coefficients. Let \[ I_p(t):=a\cos(\omega t)+b\sin(\omega t). \] Then \begin{align*} I_p'(t) &=\omega(-a\sin(\omega t)+b\cos(\omega t)),\\ I_p''(t) &=\omega^2(-a\cos(\omega t)-b\sin(\omega t)). \end{align*} Substitute them into \eqref{eq:periodicdiffeq} we have
\begin{align*}
LI_p''(t)+RI_p'(t)+\frac{1}{C}I_p(t)
&\Step{1C}{ =L(-a\omega^2\cos(\omega t)-b\omega^2\sin(\omega t))}\\
&\Step{2C}{ +R(-a\sin(\omega t)+b\omega\cos(\omega t))}\\
&\Step{3C}{ +\frac{1}{C}(a\cos(\omega t)+b\sin(\omega t))}\\
&\Step{4C}{ =(L\omega^2(-a)+R\omega b+a/C)\cos(\omega t)}\\
&\Step{5C}{ +(L\omega^2(-b)+R\omega(-a)+b/C)\sin(\omega t)}.\\
\end{align*}
Then we collect the sine and cosine terms and equate them to $0$ and $E_0\omega\cos(\omega t)$, \begin{align*} L\omega^2(-a)+R\omega b+a/C &=E_0\omega,\\ L\omega^2(-b)+R\omega(-a)+b/C &=0. \end{align*} The solution of this system is \[ a=\frac{-E_0S}{R^2+S^2}, \qquad b=\frac{E_0R}{R^2+S^2}, \] where \[ S=\omega L-\frac{1}{\omega C}. \] Here we may assume that $R\gt 0$, so $R^2+S^2\gt 0$.
Hence \[ I_p(t)=\frac{E_0}{\sqrt{R^2+S^2}}\left(\frac{R}{\sqrt{R^2+S^2}}\sin(\omega t)-\frac{S}{\sqrt{R^2+S^2}}\cos(\omega t)\right). \] Here we can transform the first term further if we use that \[ \frac{A\sin(x)-B\cos(x)}{\sqrt{A^2+B^2}}=\sin(x-\delta),\qquad(A,B\gt 0), \] where $\tan(\delta)=\frac{B}{A} $.
Applying to our case \[ I_p(t)=I_0\sin(\omega t-\delta), \] where \[ I_0=\frac{E_0}{\sqrt{R^2+S^2}}, \] and \[ \tan(\delta)=\frac{S}{R}. \]
