A heart pacemaker consists of a switch, a battery of constant voltage E0 , 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 E across the heart satisfies the linear differential equation

\f[\frac{dE}{dt}=-\frac{1}{RC}E\f]

Solve the DE, subject to E(4) =E 0 .

I’ll start attacking this problem by observing that the given differential equation is separable. So i’ll do the separation as follows:

\f[\frac{dE}{E}=-\frac{1}{RC}dt\f] .

I’ll then integrate both sides of the equation as follows:

\f[\int \frac{dE}{E}=-\int \frac{1}{RC}dt\f]

On integration we get:

\f[ln E=-\frac{t}{RC}+k\f]

Here , k is a constant of integration. I’ll then make E(t) the subject of the formula by introducing an exponential on both sides, so that:

\f[E(t)=e^{-\frac{t}{RC}+k}\f]

The exponential on the right side of the equation can be broken down into two terms using the laws of indices as follows:

\f[e^{-\frac{t}{RC}+k}=e^{-\frac{t}{RC}}*e^{k}\f]

But ek is just a constant. Therefore ,we can rewrite the above equation as follows:

\f[e^{-\frac{t}{RC}}*e^{k}=Ae^{-\frac{t}{RC}}\f]

Here,

\f[e^{k}=A\f]

The expression for E(t) then becomes;

\f[E(t)=Ae^{-\frac{t}{RC}}\f]

We are given the following boundary condition: E(4) =E 0. Plugging this into E(t),

\f[E_{0}=Ae^{-\frac{4}{RC}}\f]

Solving for A we get:

\f[A=E_{0}e^{\frac{4}{RC}}\f]

Therefore E(t) becomes:

\f[E(t)=E_{0}e^{\frac{4}{RC}}*e^{-\frac{t}{RC}}\f]

Merging the two exponential terms ,we get the following:

\f[E(t)=E_{0}e^{-\frac{1}{RC}(t-4)}\f]

That is the solution for E(t) in terms of the given constants and time t. The negative sign on the exponential term indicates a ‘decay’ in the voltage when the capacitor is discharging.