Question

Heart Pacemaker A heart pacemaker consists of a switch, a battery of constant voltage $E_0,$,a capacitor with constant capacitance C,and the heart as a resistor with constant resistance R.When the switch is closed, the capacitor charges; when the switch is open, the capacitor discharges, sending an electrical stimulus to the heart. During the time the heart is being stimulated, the voltage Eacross the heart satisfies the linear differential equation.

$\frac{dE}{dt}=-\frac{1}{RC}E.$

Solve the DE, subject to$E\left(4\right)=E_0.$

Short answer

$E=E_0{e}^{-(t-4)/RC}$

Step-by-step solution

Given Information

I'm going to start by pointing out that the given differential equation is separable. So here's how I'll separate them:

$\frac{dE}{E}=-\frac{1}{RC}dt$

Solve the problem by separating the variables.

$$\begin{gathered} \frac{1}{E}dE=-\frac{1}{RC}dt \\ \int \frac{1}{E}dE=-\int \frac{1}{RC}dt \\ \ln E=-\frac{1}{RC}t+c \\ E=c{e}^{-\frac{1}{RC}t} \end{gathered}$$

Substitute the initial condition

$$\begin{gathered} E_0=c{e}^{-\frac{1}{RC}\left(4\right)} \\ c=E_0{e}^{\frac{4}{RC}} \end{gathered}$$

Substitute C value,

$$\begin{gathered} E=\left(E_0{e}^{\frac{4}{RC}}\right)e^{-\frac{1}{RC}t} \\ E=E_0{e}^{-(t-4)/RC} \end{gathered}$$

In terms of the given constants and time t, this is the answer for When the capacitor discharges, the negative sign on the exponential term implies a 'decay' in the voltage.

Hence the solution is$E=E_0{e}^{-(t-4)/RC}$