G Solving the Feedback Equation

This appendix derives the solution to the feedback equation (Eqn. (10.1)) presented in Lecture 10. The derivation uses separation of variables and a standard integral. You are not expected to reproduce this derivation, but understanding the approach may help you see where the result comes from.

G.1 Setting Up the Integral

The feedback equation is:

\[\frac{\mathrm{d}[\mathrm{I}]}{\mathrm{d}t} = \nu_\mathrm{f} + \phi[\mathrm{I}]\]

We separate variables by dividing both sides by \(\nu_\mathrm{f} + \phi[\mathrm{I}]\) and multiplying by \(\mathrm{d}t\):

\[\frac{\mathrm{d}[\mathrm{I}]}{\nu_\mathrm{f} + \phi[\mathrm{I}]} = \mathrm{d}t\]

Integrating both sides — the left from \([\mathrm{I}]_0\) to \([\mathrm{I}]_t\), the right from \(0\) to \(t\):

\[\int_{[\mathrm{I}]_0}^{[\mathrm{I}]_t} \frac{\mathrm{d}[\mathrm{I}]}{\nu_\mathrm{f} + \phi[\mathrm{I}]} = \int_0^t \mathrm{d}t\]

G.2 Evaluating the Left-Hand Side

The left-hand integral has the form \(\int \mathrm{d}x/(a + bx)\). To evaluate this, we substitute \(u = a + bx\), so that \(\mathrm{d}u = b\,\mathrm{d}x\) and \(\mathrm{d}x = \mathrm{d}u/b\):

\[\int \frac{\mathrm{d}x}{a + bx} = \frac{1}{b}\int \frac{\mathrm{d}u}{u} = \frac{1}{b}\ln(u) = \frac{1}{b}\ln(a + bx)\]

With \(a = \nu_\mathrm{f}\) and \(b = \phi\), the left-hand side becomes:

\[\frac{1}{\phi}\Big[\ln\!\left(\nu_\mathrm{f} + \phi[\mathrm{I}]\right)\Big]_{[\mathrm{I}]_0}^{[\mathrm{I}]_t} = \frac{1}{\phi}\ln\!\left(\frac{\nu_\mathrm{f} + \phi[\mathrm{I}]_t}{\nu_\mathrm{f} + \phi[\mathrm{I}]_0}\right)\]

G.3 Solving for \([\mathrm{I}]_t\)

Setting \([\mathrm{I}]_0 = 0\) and equating with the right-hand side:

\[\frac{1}{\phi}\ln\!\left(\frac{\nu_\mathrm{f} + \phi[\mathrm{I}]_t}{\nu_\mathrm{f}}\right) = t\]

Multiplying both sides by \(\phi\) and exponentiating:

\[\frac{\nu_\mathrm{f} + \phi[\mathrm{I}]_t}{\nu_\mathrm{f}} = \mathrm{e}^{\phi t}\]

Rearranging:

\[1 + \frac{\phi[\mathrm{I}]_t}{\nu_\mathrm{f}} = \mathrm{e}^{\phi t}\]

\[[\mathrm{I}]_t = \frac{\nu_\mathrm{f}}{\phi}\left(\mathrm{e}^{\phi t} - 1\right)\]

This is Eqn. (10.2). The result is valid for any value of \(\phi\); the sign of \(\phi\) determines whether \([\mathrm{I}]\) approaches a steady state (\(\phi < 0\)) or grows exponentially (\(\phi > 0\)).