Electric Circuits 3

Problem 3.

Determine the particular solution $I_p(t)$ of the equation ($L,R,C,\gamma\gt 0$) $$LI''(t)+RI'(t)+\frac{1}{C}I(t)=E_0e^{-\gamma t}.$$

Show Solution

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

$$\begin{align*} LI_p''(t)+RI_p'(t)+\frac{1}{C}I_p(t) &\Step{1A}{ =L(-2A\gamma e^{-\gamma t}+At\gamma^2e^{-\gamma t})}\\ &\Step{2A}{ +R(Ae^{-\gamma t}-At\gamma e^{-\gamma t})+\frac{Ate^{-\gamma t}}{C}}\\ &\Step{3A}{ =Ate^{-\gamma t}\left(L\gamma^2-R\gamma+\frac{1}{C}\right)}\\ &\Step{4A}{ +e^{-\gamma t} A(2L(-\gamma)+R)}\\ &\Step{5A}{ =Ate^{-\gamma t}\underbrace{\left(L\gamma^2-R\gamma+\frac{1}{C}\right)}_{=0,\text{ by assumption}} }\\ &\Step{6A}{ +e^{-\gamma t}A\underbrace{(-2L\gamma+R)}_{\neq 0, \text{ by assumption}} }\\ &\Step{7A}{ =e^{-\gamma t}A(-2L\gamma+R) }\\ &\Step{8A}{ =E_0e^{-\gamma t}. }\\ \end{align*}$$

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 $$I_p(t)=\frac{E_0}{2L}t^2e^{-\gamma t}.$$