Recall that a rate equation relates a quantity to its rate of variation. Our goal in solving a rate equation is to relate the quantity directly to the independent variable. Since a rate of variation is a derivative function, ultimately solving a rate equation involves recognizing and reversing the differentiation. When we can “separate” the variables involved in a rate function to either side of a rate equation, on one side we can reverse the differentiation by reversing the chain rule.
Example 15.3.1. A first example.
Consider the rate equation
\begin{equation*}
r(t) = \frac{t^2 + 1}{2 q(t) + 1} \text{.}
\end{equation*}
Let’s simply write \(q\) instead of \(q(t)\text{.}\) But let’s also replace \(r(t)\) with \(dq/dt\text{,}\) since \(dq/dt\) is the rate of variation of \(q\text{.}\) Finally, we’ll also do a little algebraic rearrangement.
\begin{gather}
r(t) = \frac{t^2 + 1}{2 q(t) + 1} \notag\\
\dqdt = \frac{t^2 + 1}{2 q + 1} \notag\\
(2 q + 1) \dqdt = t^2 + 1\tag{†}\\
2 q \cdot \dqdt + \dqdt = t^2 + 1 \text{.}\notag
\end{gather}
The first term on the left now looks the type of result we would obtain when applying the chain rule during an implicit differentiation computation:
\begin{align*}
\text{formula in } q \amp = \text{formula in } t \\
\amp \downarrow d/dt \text{ ?} \\
(2 q + 1) \dqdt \amp = t^2 + 1 \text{.}
\end{align*}
Our goal is to relate quantity \(q\) to the independent variable \(t\) directly, and if we could reverse the process above and return to an equality involving \(q\) and \(t\) then we could attempt to isolate \(q\) to achieve our goal.
To reverse a suspected differentiation process, let us
antidifferentiate both sides of our re-arranged rate equation. First, we’ll tackle the right-hand side of
(†), since it is straight-forward:
\begin{equation*}
\ccmiint{t^2 + 1}{t} = \frac{t^3}{3} + t + C \text{.}
\end{equation*}
For now, let us choose the specific antiderivative with \(C = 0\text{.}\) By definition of antiderivative, our calculation above says
\begin{equation*}
t^2 + 1 = \opddt \left( \frac{t^3}{3} + t \right) \text{.}
\end{equation*}
Antidifferentiating the left-hand side of
(†), involves computing
\begin{equation*}
\ccmiint{(2 q + 1) \dqdt}{t} \text{.}
\end{equation*}
But this looks like an integral involving the method of substitution, just with \(q\) in place of \(u\text{:}\)
\begin{equation*}
\ccmiint{(2 q + 1) \dqdt}{t} = \ccmiint{2 q + 1}{q} = q^2 + q + C \text{.}
\end{equation*}
Again, let us choose the specific antiderivative with \(C = 0\text{.}\) Then our calculation above says that
\begin{equation*}
(2 q + 1) \dqdt = \opddt (q^2 + q) \text{.}
\end{equation*}
Combining these separate expressions for the two sides of
(†) back into an equality, we have
\begin{gather*}
(2 q + 1) \dqdt = t^2 + 1 \\
\opddt (q^2 + q) = \opddt \left( \frac{t^3}{3} + t \right) \text{.}
\end{gather*}
However,
Corollary 13.4.7 states that two derivatives can only be equal when the two functions being differentiated are vertical shifts of one another. In other words, we must have
\begin{equation*}
q^2 + q = \frac{t^3}{3} + t + K
\end{equation*}
for some unknown constant \(K\text{.}\)
We now at least have obtained an implicit expression relating
\(q\) to
\(t\text{.}\) It’s not always possible to isolate
\(q\text{,}\) but in this case we can use the
Quadratic formula to obtain a “lower” and an “upper” solution:
\begin{gather*}
q^2 + q = \frac{t^3}{3} + t + K\\
3 q^2 + 3 q = t^3 + 3 t + K \\
3 q^2 + 3 q - (t^3 + 3 t + K) = 0
\end{gather*}
\begin{align*}
q(t) \amp = \frac{-3 - \sqrt{9 + 12 (t^3 + 3 t + K)}}{6}
\amp
q(t) \amp = \frac{-3 + \sqrt{9 + 12 (t^3 + 3 t + K)}}{6}\\
\amp = \frac{-3 - \sqrt{K + 12 (t^3 + 3 t)}}{6} \text{,}
\amp
\amp = \frac{-3 + \sqrt{K + 12 (t^3 + 3 t)}}{6} \text{.}
\end{align*}
Notice that we have replaced \(9 + 12 K\) with just \(K\text{,}\) since if \(K\) is an arbitrary value, then so is \(9 - 12 K\text{.}\)
Our two classes of solutions both involve the parameter \(K\text{,}\) which would have to be determined using a known or desired initial value. For example, if we want a solution of the “upper” type (with the plus sign before the square root) that satisfies \(q(0) = 2\text{,}\) we could solve for \(K\text{:}\)
\begin{gather*}
2 = \frac{-3 + \sqrt{K + 12 (0^3 + 3 \cdot 0)}}{6} \\
15 = \sqrt{K} \\
225 = K \text{.}
\end{gather*}
So the solution that satisfies \(q(0) = 2\) is
\begin{equation*}
q(t) = \frac{-3 + \sqrt{225 + 12 (t^3 + 3 t)}}{6} \text{.}
\end{equation*}
Note that there is no solution of the “lower” type (with the minus sign before the square root) that satisfies \(q(0) = 2\text{,}\) since that would require \(\sqrt{K}\) to be a negative value.
Procedure 15.3.3. Solving a separable rate equation.
To solve
\begin{equation*}
m(q) \cdot \dqdt = n(t) \text{,}
\end{equation*}
integrate both sides:
\begin{gather*}
m(q) \cdot \dqdt = n(t) \\
\ccmiint{m(q) \cdot \dqdt}{t} = \ccmiint{n(t)}{t} \\
\ccmiint{m(q)}{q} = \ccmiint{n(t)}{t} \\
M(q) = N(t) + C \text{,}
\end{gather*}
where \(M(q)\) is an antiderivative for \(m(q)\) (with respect to \(q\)) and \(N(t)\) is an antiderivative for \(n(t)\) (with respect to \(t\)). Notice that we have introduced an arbitrary constant of integration on one side only.
Example 15.3.4. Solving exponential decay.
Let’s once again revisit
Example 4.2.3, which involved the rate equation
\begin{equation*}
\frac{r(t)}{q(t)} = - k \text{.}
\end{equation*}
In
Example 12.5.8 we verified that an horizontally-transformed exponential function can be a solution to this type of rate equation. Now we’ll solve this equation to verify that exponential functions are the
only possible solutions.
\begin{gather*}
\frac{1}{q} \cdot \dqdt = -k \\
\ccmiint{\frac{1}{q} \cdot \dqdt}{t} = \ccmiint{-k}{t} \\
\ccmiint{\frac{1}{q}}{q} = \ccmiint{-k}{t} \\
\ln\abs{q} = -k t + C \text{.}
\end{gather*}
To isolate \(q\text{,}\) we undo the logarithm by taking the exponential of each side.
\begin{gather*}
\exp\bbrac{\ln\abs{q}} = e^{-k t + C} \\
\abs{q} = e^C \cdot e^{-k t} \text{.}
\end{gather*}
Now, the exponentials on the left are always positive, but the \(q\) inside the absolute value brackets could be positive or negative, so again we really have a “lower” and “upper” solution
\begin{align*}
q \amp = e^C \cdot e^{-k t} \amp q \amp = - e^C \cdot e^{-k t}
\end{align*}
However, as \(C\) is an arbitrary constant, together \(\pm e^C\) can take on any non-zero value, so we unify these two solutions by letting \(A\) represent either of \(\pm e^C\text{:}\)
\begin{equation*}
q(t) = A e^{-k t} \text{.}
\end{equation*}
Notice that \(A\) represents the initial value \(q(0)\text{,}\) since \(e^0 = 1\text{.}\)
One last thing to consider is that if \(A = \pm e^C\text{,}\) then \(A = 0\) is not possible. And that’s good, because \(q = 0\) would not work in
\begin{equation*}
\frac{r(t)}{q(t)} = - k \text{.}
\end{equation*}
However, if we rearrange this to
\begin{equation*}
r(t) = - k q(t) \text{,}
\end{equation*}
then \(q(t) = 0\) would in fact be a solution, since then \(q'(t) = 0\) also, so
\begin{align*}
\text{LHS} \amp = r(t) \amp \text{RHS} \amp = -k q(t) \\
\amp = q'(t) \amp \text{RHS} \amp = -k \cdot 0 \\
\amp = 0 \amp \text{RHS} \amp = 0 \text{.}
\end{align*}
Let’s record the pattern of the solution to the previous example.
Pattern 15.3.5. Solution to proportional rate equation.
If \(q(t)\) is a varying quantity so that its rate of variation \(r(t)\) is always proportional to the amount \(q(t)\) according to the rate equation
\begin{equation*}
r(t) = k q(t) \text{,}
\end{equation*}
then \(q(t)\) is the exponential function
\begin{equation*}
q(t) = q_0 e^{k t} \text{,}
\end{equation*}
where \(q_0 = q(0)\) is the initial quantity.
Example 15.3.6.
\begin{equation*}
r(t) = \frac{q(t) \cdot \bbrac{1 - q(t)}}{t + 1} \text{.}
\end{equation*}
This is somewhat similar to
Example 15.3.1, except this time the solution will involve logarithms, because after separating variables we will have reciprocal functions on both sides.
\begin{equation*}
\frac{1}{q (1 - q)} \cdot \dqdt = \frac{1}{t + 1} \text{.}
\end{equation*}
Because of the product in the denominator, the left-hand side looks like something you get after obtaining a common denominator. In fact, combining
\begin{equation*}
\frac{1}{q} + \frac{1}{1 - q}
\end{equation*}
into a single fraction leads to our left-hand side above (without the \(dq/dt\)).
\begin{gather*}
\left(\frac{1}{q} + \frac{1}{1 - q}\right) \cdot \dqdt = \frac{1}{t + 1} \\
\ccmiint{\frac{1}{q} + \frac{1}{1 - q}}{q} = \ccmiint{\frac{1}{t + 1}}{t}
\end{gather*}
For each of the right-hand side and the second term on the left-hand side, we could perform a substitution similarly to
Example 15.2.4, but we can probably just guess-and-check these antiderivatives using the logarithm:
\begin{equation*}
\ln\abs{q} - \ln\abs{1 - q} = \ln\abs{t + 1} + C \text{,}
\end{equation*}
where the original addition on the left has turned into subtraction to correct the extra negative we would get from the chain rule if we differentiated \(\ln\abs{1 - q}\) to check our answer.
Finally, we would like to solve for \(q\) if possible. We use the properties of the logarithm to combine the two terms on the left, and then apply the exponential function to “undo” the logarithm.
\begin{gather*}
\ln\abs{\frac{q}{1 - q}} = \ln\abs{t + 1} + C \\
\exp\left(\ln\abs{\frac{q}{1 - q}}\right) = \exp\bbrac{\ln\abs{t + 1} + C} \\
\abs{\frac{q}{1 - q}} = \exp\bbrac{\ln\abs{t + 1}} \cdot e^C \\
\abs{\frac{q}{1 - q}} = e^C \abs{t + 1} \\
\frac{q}{1 - q} = \pm e^C (t + 1)
\end{gather*}
Replace \(\pm e^C\) with arbitrary \(A\text{:}\)
\begin{gather*}
q = A (t + 1)(1 - q) \\
q = A (t + 1) - A (t + 1) q \\
q + A (t + 1) q = A (t + 1) \\
\bbrac{ 1 + A (t + 1) } q = A (t + 1) \\
q(t) = \frac{A (t + 1)}{ 1 + A (t + 1) } \text{,}
\end{gather*}
where
\(A\) is an arbitrary constant. This general solution appears different from the one previously provided in
Example 4.2.11, but this solution can be converted into that solution by substituting
\begin{equation*}
A = \frac{c}{1 -c} \text{,}
\end{equation*}
where \(c\) also represents an arbitrary constant.
