Question

Suppose an RC-series circuit has a variable resistor. If the resistance at time $\mathit{t}$is defined by $\mathit{R}\left(t\right)\mathbf{=}{\mathit{k}}_{\mathbf{1}}\mathbf{+}{\mathit{k}}_{\mathbf{2}}\mathit{t}$, where ${\mathit{k}}_{\mathbf{1}}$and ${\mathit{k}}_{\mathbf{2}}$are known positive constants, then the differential equation in (9) of Section 3.1 becomes

$\left(k_1+k_{2}t\right)\frac{\mathbf{d}\mathbf{q}}{\mathbf{d}\mathbf{t}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{C}}\mathit{q}\mathbf{=}\mathit{E}\left(t\right)\mathbf{,}$

where C is a constant. If$\mathit{E}\left(t\right)\mathbf{=}{\mathit{E}}_{\mathbf{0}}$ and $\mathit{q}\left(0\right)\mathbf{=}{\mathit{q}}_{\mathbf{0}}$, where ${\mathit{E}}_{\mathbf{0}}$and${\mathit{q}}_{\mathbf{0}}$ are constants, then show that

$\mathit{q}\left(t\right)\mathbf{=}{\mathit{E}}_{\mathbf{0}}\mathit{C}\mathbf{+}\left(q_0-E_{0}C\right){\left(\frac{k_1}{k_1+k_{2}t}\right)}^{\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{C}{\mathbf{k}}_{\mathbf{2}}$}\right.}$

Short answer

The solution is $q\left(\mathrm{t}\right)=E_{0}C+({\mathrm{q}}_0-{\mathrm{E}}_0\mathrm{C}){\left(\frac{{\mathrm{k}}_1}{{\mathrm{k}}_1+{\mathrm{k}}_2\mathrm{t}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\mathrm{Ck}}_2$}\right.}$

Step-by-step solution

 Definition of differential equation

A differential equation is an equation containing the derivatives or differentials of one or more dependent variables, with respect to one or more independent variables.

 Solve the given differential equation

The RC-series circuit with a variable resistance $R\left(t\right)$is described by the differential equation $\left(k_1+k_{2}t\right)\frac{dq}{dt}+\frac{1}{C}q=E\left(t\right)$. Rearrange it as follows:\

$\frac{dq}{dt}+\frac{1}{C\left(k_1+k_{2}t\right)}q=\frac{E\left(t\right)}{k_1+k_{2}t}.............\left(1\right)$

This differential equation (1) is a first order differential equation in the form $\frac{dq}{dt}+p\left(t\right)q-g\left(t\right)$and can be solve using integrating factor technique. Obtain the integrating factor as follows:

$$\begin{gathered} \mathrm{I}.\mathrm{F}={\mathrm{e}}^{\int \mathrm{p}\left(\mathrm{t}\right)\mathrm{dt}} \\ ={\mathrm{e}}^{\int \frac{1}{\mathrm{c}\left({\mathrm{k}}_1+{\mathrm{k}}_2\mathrm{t}\right)}\mathrm{dt}} \\ ={\mathrm{e}}^{\frac{1}{{\mathrm{Ck}}_2}\int \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1+{\mathrm{k}}_2\mathrm{t}}\mathrm{dt}} \\ ={\mathrm{e}}^{\frac{1}{{\mathrm{Ck}}_2}\mathrm{In}\left({\mathrm{k}}_1+{\mathrm{k}}_2\mathrm{t}\right)} \end{gathered}$$

Solve further as

$$\begin{gathered} \mathrm{IF}={{\mathrm{e}}^{\mathrm{In}\left({\mathrm{k}}_1+{\mathrm{k}}_2\mathrm{t}\right)}}^{\frac{1}{{\mathrm{Ck}}_2}} \\ ={\left({\mathrm{k}}_1+{\mathrm{k}}_2\mathrm{t}\right)}^{\frac{1}{{\mathrm{Ck}}_2}} \end{gathered}$$

Now, multiply both sides of our differential equation in (1) by the integrating factor, then we have

$$\begin{gathered} {\left(k_1+k_{2}t\right)}^{\frac{1}{C{k}_2}}\frac{dq}{dt}+{\left(k_1+k_{2}t\right)}^{\frac{1}{C{k}_2}}\frac{1}{C\left(k_1+k_{2}t\right)}q=\frac{E\left(t\right)}{k_1+k_{2}t}{\left(k_1+k_{2}t\right)}^{\frac{1}{C{k}_2}} \\ {\left(k_1+k_{2}t\right)}^{\frac{1}{C{k}_2}}\frac{dq}{dt}+\frac{1}{C}{\left(k_1+k_{2}t\right)}^{\frac{1-C{k}_2}{C{k}_2}}q=E\left(t\right){\left(k_1+k_{2}t\right)}^{\frac{1-C{k}_2}{C{k}_2}} \end{gathered}$$

Simplify further

$$\begin{gathered} \frac{d}{dt}{\left(k_1+k_{2}t\right)}^{\frac{1}{C{k}_2}}q=E\left(t\right){\left(k_1+k_{2}t\right)}^{\frac{1-C{k}_2}{C{k}_2}} \\ \int d{\left(k_1+k_{2}t\right)}^{\frac{1}{C{k}_2}}q=E\left(t\right)\int {\left(k_1+k_{2}t\right)}^{\frac{1-C{k}_2}{C{k}_2}} \\ {\left(k_1+k_{2}t\right)}^{\frac{1}{C{k}_2}}q=C{k}_{2}E\left(t\right){\left(k_1+k_{2}t\right)}^{\frac{1}{C{k}_2}}\times \frac{1}{k_2} \\ {\left(k_1+k_{2}t\right)}^{\frac{1}{C{k}_2}}q=CE\left(t\right){\left(k_1+k_{2}t\right)}^{\frac{1}{C{k}_2}}+h \\ \mathrm{Thus}\mathrm{q}\left(\mathrm{t}\right)={\mathrm{E}}_0\mathrm{C}+\mathrm{h}{\left(\frac{1}{{\mathrm{k}}_1+{\mathrm{k}}_2\mathrm{t}}\right)}^{{}^{\frac{1}{{\mathrm{Ck}}_2}}}\mathrm{where}\mathrm{E}\left(\mathrm{t}\right)={\mathrm{E}}_0. \end{gathered}$$

Find the value of the constant

To find the value of constant $h$, apply the point of condition $\left(q,t\right)=\left(q_0,0\right)$into equation $q\left(t\right)=E_{0}C+h{\left(\frac{1}{k_1+K_{2}t}\right)}^{\frac{1}{C{k}_2}}$, then we have

$$\begin{gathered} q_0=E_{0}C+h{\left(\frac{1}{k_1}\right)}^{\frac{1}{C{k}_2}} \\ {q}_0-E_{0}C=h{\left(\frac{1}{k_1}\right)}^{\frac{1}{C{k}_2}} \\ \mathrm{Thus},h=\left(q_0-E_{0}C\right)k^{\left(\frac{1}{C{k}_2}\right)} \\ \mathrm{Substitute}\mathrm{h}=\left({\mathrm{q}}_0-{\mathrm{E}}_0\mathrm{C}\right){\mathrm{k}}^{\left(\frac{1}{{\mathrm{Ck}}_2}\right)}\mathrm{for}\mathrm{h}\mathrm{into}\mathrm{equation}\mathrm{q}\left(\mathrm{t}\right)={\mathrm{E}}_0\mathrm{C}+\mathrm{h}{\left(\frac{1}{{\mathrm{k}}_1+{\mathrm{k}}_2\mathrm{t}}\right)}^{\frac{1}{{\mathrm{Ck}}_2}}, \\ \mathrm{q}\left(\mathrm{t}\right)={\mathrm{E}}_0\mathrm{C}+\left({\mathrm{q}}_0-{\mathrm{E}}_0\mathrm{C}\right){\mathrm{k}}^{\left(\frac{1}{{\mathrm{Ck}}_2}\right)}{\left(\frac{1}{{\mathrm{k}}_1+{\mathrm{k}}_2\mathrm{t}}\right)}^{\frac{1}{{\mathrm{Ck}}_2}} \\ \mathrm{q}\left(\mathrm{t}\right)={\mathrm{E}}_0\mathrm{C}+\left({\mathrm{q}}_0-{\mathrm{E}}_0\mathrm{C}\right){\left(\frac{{\mathrm{k}}_1}{{\mathrm{k}}_1+{\mathrm{k}}_2\mathrm{t}}\right)}^{\frac{1}{{\mathrm{Ck}}_2}} \\ \mathrm{Hence}\mathrm{it}\mathrm{is}\mathrm{proved}\mathrm{that}\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{q}\left(\mathrm{t}\right)={\mathrm{E}}_0\mathrm{C}+\left({\mathrm{q}}_0-{\mathrm{E}}_0\mathrm{C}\right){\left(\frac{{\mathrm{k}}_1}{{\mathrm{k}}_1+{\mathrm{k}}_2\mathrm{t}}\right)}^{\frac{1}{{\mathrm{Ck}}_2}} \end{gathered}$$