Electronic – Source-free, under-damped, parallel RLC with 2 intial conditions

Just to be complete:

^{simulate this circuit – Schematic created using CircuitLab}

I gather that's the circuit. And you'd like an expression for \$V_{\left(t\right)}\$ and \$I_{L\left(t\right)}\$ that handles the initial conditions where \$V_{\left(0\right)}\$ and \$I_{L\left(0\right)}\$ may be non-zero and are otherwise known at \$t=0\$.

You also want to focus on the case that is underdamped.

I hope you won't mind if I start at the basics and take this through, carefully. A lot of it will just be repetition of the obvious. I apologize for that.

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The basic nodal equation is (treating all currents pointing downwards as positive):

$$\frac{V_{\left(t\right)}}{R}+\frac{1}{L}\int V_{\left(t\right)}\:\text{d}t+C\frac{\text{d}V_{\left(t\right)}}{\text{d}t}=0\:\text{A}$$

Taking the derivative and dividing through by \$C\$ (had I added \$I_{L\left(0\right)}\$ above, it would disappear now, anyway):

$$\begin{align*}\frac{\text{d}^2V_{\left(t\right)}}{\text{d}t^2}+\frac{1}{R\:C}\frac{\text{d} V_{\left(t\right)}}{\text{d}t}+\frac{V_{\left(t\right)}}{L\:C}&=0\:\text{A}\end{align*}$$

A standard 2nd order ODE where the obvious proposed solution is \$V_{\left(t\right)}=A e^{s t}\$. By substitution, we find the characteristic equation is:

$$s^2+\frac{s}{R\:C}+\frac{1}{L\:C}=0$$

Solving with the standard quadratic solution equation, \$\frac{-b\pm\sqrt{b^2-4\:a\:c}}{2\:a}\$, we find that it is convenient (from a cursory examination of the results) to define:

$$\begin{align*}\alpha&=\frac{1}{2\:R\:C}&\omega_0&=\frac{1}{\sqrt{L\:C}}\end{align*}$$

The solutions are then:

$$\begin{align*}s_1&=-\alpha+\sqrt{\alpha^2-\omega_0^2}&s_2&=-\alpha-\sqrt{\alpha^2-\omega_0^2}\end{align*}$$

At this point, there are three possibilities to consider. One is the critically damped case where \$\alpha = \omega_0\$ and therefore \$s_1=s_2\$ and they are both real-valued. Another is the overdamped case where \$\alpha > \omega_0\$ and therefore \$s_1\ne s_2\$ but they are both real-valued, again. The final case is the underdamped case where \$\alpha < \omega_0\$ and \$s_1\$ and \$s_1\$ are both complex-valued and are conjugates of each other.

This last case is the one you want to address.

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In the underdamped case it is again convenient to flip the sign inside the square-root and define another real-valued variable, the damped frequency:

$$\begin{align*}\omega_d&=\sqrt{\omega_0^2-\alpha^2}&&=\sqrt{-1}\cdot\sqrt{\alpha^2-\omega_0^2}\\\\&\therefore\\\\s_1&=-\alpha+j\:\omega_d&s_2&=-\alpha-j\:\omega_d\end{align*}$$

Now the general solution (in the underdamped case there are two conjugate answers, so both are valid and should be included) is:

$$\begin{align*}V_{\left(t\right)}&=A_1\:e^{s_1\:t}+A_2\:e^{s_2\:t}\\\\ &=e^{-\alpha\:t}\left[\:\left(A_1+A_2\right)\operatorname{cos}\left(\omega_d\:t\right)+j\:\left(A_1-A_2\right)\operatorname{sin}\left(\omega_d\:t\right)\:\right]\\\\ \text{setting: } B_1&=A_1+A_2\\B_2&=j\left(A_1-A_2\right)\\\\ &=e^{-\alpha\:t}\left[\:B_1\operatorname{cos}\left(\omega_d\:t\right)+B_2\operatorname{sin}\left(\omega_d\:t\right)\:\right]\end{align*}$$

Clearly,

$$\begin{align*}V_{\left(0\right)}&=B_1\end{align*}$$

The derivative is,

$$\begin{align*} \frac{\text{d}V_{\left(t\right)}}{\text{d}t}&=e^{-\alpha\:t}\left[\:\left(\omega_d\:B_2-\alpha\:B_1\right)\operatorname{cos}\left(\omega_d\:t\right)-\left(\omega_d\:B_1+\alpha\:B_2\right)\operatorname{sin}\left(\omega_d\:t\right)\:\right] \end{align*}$$

Clearly,

$$\begin{align*} \frac{\text{d}V_{\left(0\right)}}{\text{d}t}&=\omega_d\:B_2-\alpha\:B_1 \end{align*}$$

You know that:

$$\begin{align*}\frac{V_{\left(0\right)}}{R}+C\frac{\text{d}V_{\left(0\right)}}{\text{d} t}+I_{L\left(0\right)}&=0\:\text{A}\\\\\therefore\\\\\frac{V_{\left(0\right)}}{R\:C}+\omega_d\:B_2-\alpha\:B_1+\frac{I_{L\left(0\right)}}{C}&=0\:\text{A}\\\\\omega_d\:B_2-\alpha\:B_1&=-\frac{V_{\left(0\right)}}{R\:C}-\frac{I_{L\left(0\right)}}{C}\\\\\omega_d\:B_2-\alpha\:V_{\left(0\right)}&=-\frac{V_{\left(0\right)}}{R\:C}-\frac{I_{L\left(0\right)}}{C}\\\\\therefore\end{align*}$$$$\begin{align*}B_1 &=V_{\left(0\right)}&B_2&=\frac{V_{\left(0\right)}\left(\alpha-\frac{1}{R\:C}\right)-\frac{I_{L\left(0\right)}}{C}}{\omega_d}=-\frac{\alpha\:V_{\left(0\right)}+\frac{I_{L\left(0\right)}}{C}}{\omega_d}\\\\&&&\text{note above: }\tfrac{1}{R\:C}=2\alpha\\\\&&\text{If }V_{\left(0\right)}\ne 0\:\text{V then }\\\\&&\eta&=-\frac{\alpha+\frac{I_{L\left(0\right)}}{V_{\left(0\right)}\:C}}{\omega_d}\\\\&&B_2&=\eta\: V_{\left(0\right)}\end{align*}$$

And that solves the time-dependent voltage equation with the use of initial conditions.

$$\begin{align*}V_{\left(t\right)}=\left\{\begin{array}{l}V_{\left(0\right)}\ne 0\:\text{V},&&V_{\left(0\right)}\:e^{-\alpha\:t}\big[\operatorname{cos}\left(\omega_d\:t\right)+\eta\:\operatorname{sin}\left(\omega_d\:t\right)\big]\\&&V_{\left(0\right)}\:e^{-\alpha\:t}\:\sqrt{1+\eta^2}\operatorname{sin}\left(\omega_d\:t+\operatorname{tan}^{-1}\left(\frac{1}{\eta}\right)\right)\\\\V_{\left(0\right)}= 0\:\text{V},&&-\frac{I_{L\left(0\right)}}{\omega_d\,C}\:e^{-\alpha\:t}\:\operatorname{sin}\left(\omega_d\:t\right)\end{array}\right.\end{align*}$$

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At this point, the remaining question is the time-dependent current equation for the inductor.

$$\begin{align*} I_{L\left(t\right)}&=\frac{1}{L}\int V_{\left(t\right)}\:\text{d} t\\\\ &=\frac{V_{\left(0\right)}}{L}\left[\int e^{-\alpha\:t}\operatorname{cos}\left(\omega_d\:t\right)\:\text{d} t+\eta \int e^{-\alpha\:t}\operatorname{sin}\left(\omega_d\:t\right)\:\text{d} t\right]\\\\ &=\frac{V_{\left(0\right)}}{L}\cdot\frac{-e^{-\alpha\:t}}{\alpha^2+\omega_d^2}\bigg[\left(\alpha\operatorname{cos}\left(\omega_d\:t\right)-\omega_d\operatorname{sin}\left(\omega_d\:t\right)\right)+\eta\cdot\left(\alpha\operatorname{sin}\left(\omega_d\:t\right)+\omega_d\operatorname{cos}\left(\omega_d\:t\right)\right) \bigg]+D_0\\\\ &=\frac{V_{\left(0\right)}}{L}\cdot\frac{-e^{-\alpha\:t}}{\alpha^2+\omega_d^2}\bigg[\left(\alpha+\eta\:\omega_d\right)\operatorname{cos}\left(\omega_d\:t\right)+\left(\eta\:\alpha-\omega_d\right)\operatorname{sin}\left(\omega_d\:t\right)\bigg]+D_0 \end{align*}$$

So,

$$\begin{align*} I_{L\left(0\right)}&=-\frac{V_{\left(0\right)}}{L}\frac{\alpha+\eta\:\omega_d}{\alpha^2+\omega_d^2}+D_0\\\\\therefore\\\\D_0&=I_{L\left(0\right)}+\frac{V_{\left(0\right)}}{L}\frac{\alpha+\eta\:\omega_d}{\alpha^2+\omega_d^2}\end{align*}$$

Resulting in,

$$\begin{align*}I_{L\left(t\right)}=\left\{\begin{array}{l}V_{\left(0\right)}\ne 0\:\text{V},&&I_{L\left(0\right)}+\frac{V_{\left(0\right)}}{L}\left[\frac{\alpha+\eta\:\omega_d-e^{-\alpha\:t}\big[\left(\alpha+\eta\:\omega_d\right)\operatorname{cos}\left(\omega_d\:t\right)+\left(\eta\:\alpha-\omega_d\right)\operatorname{sin}\left(\omega_d\:t\right)\big]}{\alpha^2+\omega_d^2}\right]\\&&I_{L\left(0\right)}+\frac{V_{\left(0\right)}}{L}\left[\frac{\alpha+\eta\:\omega_d}{\alpha^2+w_d^2}-e^{-\alpha\:t}\sqrt{1+\eta^2}\frac{\operatorname{sin}\left(\omega_d\:t+\operatorname{tan}^{-1}\left[\frac{\alpha+\eta\:\omega_d}{\eta\:\alpha-\omega_d}\right]\right)}{\sqrt{\alpha^2+w_d^2}}\right]\\\\V_{\left(0\right)}= 0\:\text{V},&&I_{L\left(0\right)}+\frac{B_2}{L}\left[\frac{\omega_d-e^{-\alpha\:t}\big[\omega_d\operatorname{cos}\left(\omega_d\:t\right)+\alpha\operatorname{sin}\left(\omega_d\:t\right)\big]}{\alpha^2+\omega_d^2}\right]\\&&I_{L\left(0\right)}+\frac{B_2}{L}\left[\frac{\omega_d}{\alpha^2+w_d^2}-e^{-\alpha\:t}\frac{\operatorname{sin}\left(\omega_d\:t+\operatorname{tan}^{-1}\left[\frac{\omega_d}{\alpha}\right]\right)}{\sqrt{\alpha^2+w_d^2}}\right]\end{array}\right.\end{align*}$$