Problem 3.
Determine the particular solution $I_p(t)$ of the equation ($L,R,C,\gamma\gt 0$)
\[
LI''(t)+RI'(t)+\frac{1}{C}I(t)=E_0e^{-\gamma t}.
\]
We seek a particular solution by the method of undetermined coefficients. Let
\[
I_p(t):=Ae^{-\gamma t}.
\]
Substituting into the equation we have
\[
A\left(L(-\gamma)^2+R(-\gamma)+\frac{1}{C} \right)=E_0.
\]
Case (a). $\boldsymbol{L\gamma^2-R\gamma+\frac{1}{C}\neq 0}$.
Then \[ A=\frac{E_0}{L\gamma^2-R\gamma+\frac{1}{C}}, \] and \[ I_p(t)=\frac{E_0}{L\gamma^2-R\gamma+\frac{1}{C}}e^{-\gamma t}. \] Case (b). $\boldsymbol{L\gamma^2-R\gamma+\frac{1}{C}= 0}$.
(i) $\,\boldsymbol{2L(-\gamma)+R\neq 0}$.
In this case we seek a particular solution in the form \[ I_p(t):=Ate^{-\gamma t}. \] Substituting it into the equation we have
Then \[ A=\frac{E_0}{-2L\gamma+R} \] and \[ I_p(t)=\frac{E_0}{-2L\gamma+R}te^{-\gamma t}. \] (ii) $\,\boldsymbol{2L(-\gamma)+R= 0}$.
In this case we seek a particular solution in the form \[ I_p(t):=At^2e^{-\gamma t}. \] Substituting it into the equation we have
Then \[ A=\frac{E_0}{2L} \] and \[ I_p(t)=\frac{E_0}{2L}t^2e^{-\gamma t}. \]
Then \[ A=\frac{E_0}{L\gamma^2-R\gamma+\frac{1}{C}}, \] and \[ I_p(t)=\frac{E_0}{L\gamma^2-R\gamma+\frac{1}{C}}e^{-\gamma t}. \] Case (b). $\boldsymbol{L\gamma^2-R\gamma+\frac{1}{C}= 0}$.
(i) $\,\boldsymbol{2L(-\gamma)+R\neq 0}$.
In this case we seek a particular solution in the form \[ I_p(t):=Ate^{-\gamma t}. \] Substituting it into the equation we have
\begin{align*}
LI_p''(t)+RI_p'(t)+\frac{1}{C}I_p(t)
&\Step{1A}{ =L(-2A\gamma e^{-\gamma t}+At\gamma^2e^{-\gamma t})}\\
&\Step{2A}{ +R(Ae^{-\gamma t}-At\gamma e^{-\gamma t})+\frac{Ate^{-\gamma t}}{C}}\\
&\Step{3A}{ =Ate^{-\gamma t}\left(L\gamma^2-R\gamma+\frac{1}{C}\right)}\\
&\Step{4A}{ +e^{-\gamma t} A(2L(-\gamma)+R)}\\
&\Step{5A}{ =Ate^{-\gamma t}\underbrace{\left(L\gamma^2-R\gamma+\frac{1}{C}\right)}_{=0,\text{ by assumption}} }\\
&\Step{6A}{ +e^{-\gamma t}A\underbrace{(-2L\gamma+R)}_{\neq 0, \text{ by assumption}} }\\
&\Step{7A}{ =e^{-\gamma t}A(-2L\gamma+R) }\\
&\Step{8A}{ =E_0e^{-\gamma t}. }\\
\end{align*}
Then \[ A=\frac{E_0}{-2L\gamma+R} \] and \[ I_p(t)=\frac{E_0}{-2L\gamma+R}te^{-\gamma t}. \] (ii) $\,\boldsymbol{2L(-\gamma)+R= 0}$.
In this case we seek a particular solution in the form \[ I_p(t):=At^2e^{-\gamma t}. \] Substituting it into the equation we have
\begin{align*}
LI_p''(t)+RI_p'(t)+\frac{1}{C}I_p(t)
&\Step{1B}{ =L(2A e^{-\gamma t}-4At\gamma e^{-\gamma t}+At^2\gamma^2 e^{-\gamma t})}\\
&\Step{2B}{ +R(2At e^{-\gamma t}-At^2\gamma e^{-\gamma t})+\frac{At^2e^{-\gamma t}}{C}}\\
&\Step{3B}{ =At^2e^{-\gamma t}\left(L\gamma^2-R\gamma+\frac{1}{C}\right)}\\
&\Step{4B}{ +Ate^{-\gamma t} (-4L\gamma+2R)}\\
&\Step{5B}{ +Ae^{-\gamma t}2L}\\
&\Step{6B}{ =At^2e^{-\gamma t}\underbrace{\left(L\gamma^2-R\gamma+\frac{1}{C}\right)}_{=0,\text{ by assumption}} }\\
&\Step{7B}{+Ate^{-\gamma t} \underbrace{(-4L\gamma+2R)}_{=0,\text{ by assumption}} }\\
&\Step{8B}{+Ae^{-\gamma t}2L }\\
&\Step{9B}{=2AL e^{-\gamma t}}\\
&\Step{10B}{ =E_0e^{-\gamma t}. }\\
\end{align*}
Then \[ A=\frac{E_0}{2L} \] and \[ I_p(t)=\frac{E_0}{2L}t^2e^{-\gamma t}. \]
