## Appendix

### Solving a differential equation using an equilibrium value

In the section Another differential equation for growth and decay, we gave a method for solving differential equations of the form

\[ \dfrac{dx}{dt} = kx + m, \]where \(k, m\) are real constants and \(k \neq 0\). We now give an alternative solution method.

We have seen that \(-\dfrac{m}{k}\) is the equilibrium value of \(x\). We define a new variable \(y\) as

\[ y = x - \Bigl( - \dfrac{m}{k} \Bigr) = x + \dfrac{m}{k}. \]So \(y\) is just a translation of \(x\), and the equilibrium value of \(x=-\dfrac{m}{k}\) corresponds to \(y=0\).

We may now express the differential equation in terms of \(y\). Since \(x = y - \dfrac{m}{k}\), we have

\begin{align*} \dfrac{dy}{dt} &= \dfrac{dx}{dt} \\ &= kx + m \\ &= k\Bigl( y - \dfrac{m}{k} \Bigr) + m \\ &= ky. \end{align*}By changing variables from \(x\) to \(y\), we have obtained a simpler equation. The general solution is

\[ y = C e^{kt}, \]where \(C\) is a real constant. Substituting back for \(x\), we obtain the general solution

\[ x = C e^{kt} - \dfrac{m}{k}. \]