## 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

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

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