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Calculus for Biological Scientists

Jeff Shriner

Section 4.2 Solving Pure-Time Differential Equations

We have seen that finding the family of solutions to a pure-time differential equation \(\frac{dy}{dt} = f(t)\) amounts to answering the following question: which functions have a derivative equal to \(f(t)\text{?}\) In other words, we must find the family of antiderivatives of \(f(t)\text{.}\) In Section 4.1 we focused on verifying whether functions were solutions and sketching graphs of solutions to differential equations that were represented graphically. In this section, we will gain methods for computing a family of antiderivatives. We first give some common vocabulary and notation for representing a family of antiderivatives.

Indefinite Integral.

Given a function \(f(x)\text{,}\) the indefinite integral of \(f\) is the family of antiderivatives of \(f\text{,}\) written as
\begin{equation*} \int f(x) dx\text{.} \end{equation*}
In Example 4.1.4 we saw that the family of antiderivatives of \(4t + 1\) was \(2t^2 + t + C\text{.}\) Equivalently, we could write this as
\begin{equation*} \int (4t + 1) dt = 2t^2 + t +C\text{.} \end{equation*}
When we find an antiderivative, we will often say that we evaluate an indefinite integral. Just as the notation \(\frac{d}{dx}[\Box]\) means “take the derivative of \(\Box\) with respect to the variable \(x\)”, the notation \(\int \Box dx\) means “find a function of \(x\) whose derivative is \(\Box\text{.}\)” It is important to pay attention to the notation to know what letter is considered as the independent variable (\(dx\text{,}\) \(dt\text{,}\) etc.).

Warm-Up 4.2.1.

Use your knowledge of derivatives of basic functions to complete Table 4.2.1 of antiderivatives. For each entry, your task is to find a family of functions whose derivatives are the given function \(f\text{.}\)
Table 4.2.1. Familiar basic functions and their antiderivatives.
\(f(x)\) \(\int f(x) dx\)  
\(k\text{,}\) (\(k\) is constant)
\(x^2\)
\(x^{0.5}\)
\(x^n\text{,}\) \(n \ne -1\)
\(\frac{1}{x}\text{,}\) \(x \neq 0\)
\(\sin(x)\)
\(\cos(x)\)
\(e^x\)
\(2^x\)
\(b^x\) \((b \gt 0, b \neq 1)\)

Subsection 4.2.1 Basic Antiderivatives

Evaluating the indefinite integral of a single basic function involves reversing a single derivative rule. We summarize the basic antiderivative rules from Warm-Up 4.2.1 below.

Basic Antiderivative Rules.

  1. (Reverse Power Rule) \(\displaystyle \int x^n dx = \frac{1}{n+1}x^{n+1} + C\text{,}\) for \(n \neq -1\)
  2. \(\int x^{-1} dx = \ln|x| +C\text{,}\) for \(x \neq 0\)
  3. (Reverse Exponential Rule) \(\displaystyle \int b^x dx = \frac{1}{\ln(b)}b^x + C\text{,}\) for \(b \gt 0, b \neq 1\)
  4. (Reverse Trig Rules) \(\displaystyle \int \sin(x) dx = -\cos(x) + C\) and \(\displaystyle \int \cos(x) dx = \sin(x) + C\)
Similar to the constant multiple and sum rules for derivatives in Subsection 2.5.3, we have constant multiple and sum rules for antiderivatives that allow us to compute indefinite integrals for many functions using just these basic rules.

Constant Multiple Rule.

For any real number \(k\text{,}\)
\begin{equation*} \int kf(x) dx = k\int f(x) dx\text{.} \end{equation*}

Sum Rule.

For two functions \(f(x)\) and \(g(x)\text{,}\)
\begin{equation*} \int (f(x)+g(x)) dx = \int f(x) dx + \int g(x) dx\text{.} \end{equation*}

Example 4.2.2. Solving Basic Pure-Time Differential Equations.

Let \(\frac{ds}{dt} = e^t - 2\sin(t) + 4t^{3}\text{,}\) \(s(0)=2\text{.}\) To solve this initial value problem, we will first find the general solution by evaluating
\begin{align*} \int (e^t - 2\sin(t) + 4t^{3}) dt \amp= \int e^t dt + \int -2\sin(t) dt + \int 4t^3 dt \\ \amp= \int e^t dt - 2\int \sin(t) dt + 4\int t^3 dt\\ \amp= e^t - 2(-\cos(t)) + 4 \cdot \frac{1}{4}t^4 + C \\ \amp= e^t +2\cos(t) + t^4 + C\text{.} \end{align*}
 1  Now we use the initial value \(s(0)=2\) to find the specific solution:
\begin{equation*} 2 = e^0 + 2\cos(0) + 0^4 + C = 1 +2 + C\text{,} \end{equation*}
which implies \(C=-1\text{.}\) Thus, the specific solution to this initial value problem is \(s(t) = e^t +2\cos(t) + t^4 -1\text{.}\)

Activity 4.2.2.

Solve each initial value problem below. Practice using correct notation for evaluating indefinite integrals.
  1. \(\frac{dr}{dt}= t^{-1} - 3\sqrt{t}\text{,}\) \(r(1) = 5\)
  2. \(\frac{dp}{dt}= \cos(t) + 2\cdot 1.5^t\text{,}\) \(p(0) = \ln(1.5)\)
  3. \(\frac{dy}{dt} = e^{0.5t}\text{,}\) \(y(0) = 0\)

Subsection 4.2.2 \(u\)-Substitution

The initial value problem in Activity 4.2.2 part 3 is slightly different in nature than the others, and required one additional small step for computing the general solution. A good first guess would be that \(\int e^{0.5t}dt = e^{0.5t} + C\text{.}\) However, we can see that this is incorrect by computing that \(\frac{d}{dt}[e^{0.5t} + C] = 0.5e^{0.5t}\) (using the Chain Rule), which is not the same function that we started with under the integral. However, since it is only off by a constant multiple, we can make an easy correction to our original guess by dividing by this constant multiple to get
\begin{equation*} \int e^{0.5t}dt =\frac{1}{0.5}\cdot e^{0.5t} + C = 2 e^{0.5t} + C\text{.} \end{equation*}
The general reason this problem was more difficult was because we were integrating a composition: if \(f(t)=e^t\) and \(g(t) = 0.5t\text{,}\) then \(e^{0.5t} = f(g(t))\text{.}\) The general reason our correction worked is that the “inside function” \(g(t) = 0.5t\) is linear, and so we just needed to divide by its derivative (slope). This is a correction that will work every time in this context:
If \(\int f(x) dx = F(x) + C\text{,}\) then
\begin{equation*} \int f(ax+b) dx = \frac{1}{a} \cdot F(ax+b) + C\text{.} \end{equation*}

Activity 4.2.3.

Compute each indefinite integral below.
  1. \(\displaystyle \displaystyle \int e^{-t} dt\)
  2. \(\displaystyle \displaystyle \int (5t - 11)^{10} dt\)
  3. \(\displaystyle \displaystyle \int \frac{1}{2x - 1} dx\)
  4. \(\displaystyle \displaystyle \int \cos(-0.1\theta + 4) d\theta\)
How can we compute more complicated integrals like this when the inside function is not linear? For example, how might we compute \(\int xe^{x^2}dx\text{?}\) Similar to the case when the inside function is linear, evaluating integrals like this is related to the chain rule for taking derivatives.
Recall that the Chain Rule states that
\begin{equation*} \frac{d}{dx}[f(g(x))] = f'(g(x) \cdot g'(x)\text{.} \end{equation*}
Restating this in terms of antiderivatives this says that
\begin{equation} \int f'(g(x) \cdot g'(x) dx = f(g(x)) + C\text{.}\tag{4.2.1} \end{equation}
Equation (4.2.1) tells us that if we can view a given function as \(f'(g(x)) g'(x)\) for some appropriate choices of \(f\) and \(g\text{,}\) then we can antidifferentiate the function by reversing the Chain Rule. Note that both \(g(x)\) and \(g'(x)\) appear in the form of \(f'(g(x)) g'(x)\text{;}\) we will sometimes say that we seek to identify a function-derivative pair (\(g(x)\) and \(g'(x)\)) when trying to apply the rule in Equation (4.2.1).
If we can identify a function-derivative pair, we will introduce a new variable \(u\) to represent the function \(g(x)\text{.}\) With \(u = g(x)\text{,}\) it follows that \(\frac{du}{dx} = g'(x)\text{,}\) so that in terms of differentials 2 , \(du = g'(x)\, dx\text{.}\) Now converting the indefinite integral to a new one in terms of \(u\text{,}\) we have
\begin{equation*} \int f'(g(x)) g'(x) \, dx = \int f'(u) \,du\text{.} \end{equation*}
Provided that \(f'\) is an elementary function whose antiderivative is known, we can easily evaluate the indefinite integral in \(u\text{,}\) and then go on to determine the desired overall antiderivative of \(f'(g(x)) g'(x)\text{.}\) We call this process \(u\)-substitution, and summarize the rule as follows:
With the substitution \(u = g(x)\text{,}\)
\begin{equation*} \int f'(g(x)) g'(x) \, dx = \int f'(u) \,du = f(u) + C = f(g(x)) + C\text{.} \end{equation*}
To see \(u\)-substitution at work, we consider the following example.

Example 4.2.3.

Evaluate the indefinite integral
\begin{equation*} \int x^3 \cdot \sin (7x^4 + 3) \, dx \end{equation*}
and check the result by differentiating.
Solution.
We can make two algebraic observations regarding the integrand, \(x^3 \cdot \sin (7x^4 + 3)\text{.}\) First, \(\sin (7x^4 + 3)\) is a composite function; as such, we know we’ll need a more sophisticated approach to antidifferentiating. Second, \(x^3\) is almost the derivative of \((7x^4 + 3)\text{;}\) the only issue is a missing constant. Thus, \(x^3\) and \((7x^4 + 3)\) are nearly a function-derivative pair. Furthermore, we know the antiderivative of \(f(u) = \sin(u)\text{.}\) The combination of these observations suggests that we can evaluate the given indefinite integral by reversing the chain rule through \(u\)-substitution.
Letting \(u\) represent the inner function of the composite function \(\sin (7x^4 + 3)\text{,}\) we have \(u = 7x^4 + 3\text{,}\) and thus \(\frac{du}{dx} = 28x^3\text{.}\) In differential notation, it follows that \(du = 28x^3 \, dx\text{,}\) and thus \(x^3 \, dx = \frac{1}{28} \, du\text{.}\) The original indefinite integral may be slightly rewritten as
\begin{equation*} \int \sin (7x^4 + 3) \cdot x^3 \, dx\text{,} \end{equation*}
and so by substituting \(u\) for \(7x^4 + 3\) and \(\frac{1}{28} \, du\) for \(x^3 \, dx\text{,}\) it follows that
\begin{equation*} \int \sin (7x^4 + 3) \cdot x^3 \, dx = \int \sin(u) \cdot \frac{1}{28} \, du\text{.} \end{equation*}
Now we may evaluate the easier integral in \(u\text{,}\) and then replace \(u\) by the expression \(7x^4 + 3\text{.}\) Doing so, we find
\begin{align*} \int \sin (7x^4 + 3) \cdot x^3 \, dx &= \int \sin(u) \cdot \frac{1}{28} \, du\\ &= \frac{1}{28} \int \sin(u) \, du\\ &= \frac{1}{28} (-\cos(u)) + C\\ &= -\frac{1}{28} \cos(7x^4 + 3) + C\text{.} \end{align*}
To check our work, we observe by the Chain Rule that
\begin{equation*} \frac{d}{dx} \left[ -\frac{1}{28}\cos(7x^4 + 3) \right] = -\frac{1}{28} \cdot (-1)\sin(7x^4 + 3) \cdot 28x^3 = \sin(7x^4 + 3) \cdot x^3\text{,} \end{equation*}
which is indeed the original integrand.
The \(u\)-substitution worked because the function multiplying \(\sin (7x^4 + 3)\) was \(x^3\text{.}\) If instead that function was \(x^2\) or \(x^4\text{,}\) the substitution process would not have worked. This is one of the primary challenges of antidifferentiation: slight changes in the integrand make tremendous differences. For instance, we can use \(u\)-substitution with \(u = x^2\) and \(du = 2xdx\) to find that
\begin{align*} \int xe^{x^2} \, dx &= \int e^u \cdot \frac{1}{2} \, du\\ &= \frac{1}{2} \int e^u \, du\\ &= \frac{1}{2} e^u + C\\ &= \frac{1}{2} e^{x^2} + C\text{.} \end{align*}
However, for the similar indefinite integral
\begin{equation*} \int e^{x^2} \, dx\text{,} \end{equation*}
the \(u\)-substitution \(u = x^2\) is no longer possible because the factor of \(x\) is missing. Hence, part of the lesson of \(u\)-substitution is just how specialized the process is: it only applies to situations where, up to a missing constant, the integrand is the result of applying the Chain Rule to a different, related function.

Activity 4.2.4.

Evaluate each of the following indefinite integrals by using these steps:
  • Find two functions within the integrand that form (up to a possible missing constant) a function-derivative pair;
  • Make a substitution and convert the integral to one involving \(u\) and \(du\text{;}\)
  • Evaluate the new integral in \(u\text{;}\)
  • Convert the resulting function of \(u\) back to a function of \(x\) by using your earlier substitution;
  • Check your work by differentiating the function of \(x\text{.}\) You should come up with the integrand originally given.
  1. \(\displaystyle \int \frac{x^2}{5x^3+1} \, dx\)
  2. \(\displaystyle \int e^x \sin(e^x) \, dx\)
  3. \(\displaystyle \int \frac{\cos(\sqrt{x})}{\sqrt{x}} \, dx\)

Subsection 4.2.3 Summary

  • Question 4.2.4.

    How can we solve pure-time differential equations that are sums of basic functions?
    Answer.
    To find the general solution of a pure-time differential equation \(\frac{dy}{dt} =f(t)\text{,}\) we need to find the family of antiderivatives \(f(t)\text{,}\) \(\int f(t) dt\text{.}\) When \(f(t)\) is a single basic function, we can reverse the appropriate rule using the basic antiderivative rules. When \(f(t)\) is made up of a sum and/or constant multiple of these basic functions, we can break it up using the constant multiple rule and the sum rule.

  • Question 4.2.5.

    How can we solve more complicated pure-time differential equations?
    Answer.
    For pure-time differential equations that are defined with an appropriate “function-derivative pair”, we can use the method of \(u\)-substitution to find the general solution.

Exercises 4.2.4 Exercises

1.

Given that \(s''(t) = \sin(t) + t\text{,}\) \(s'(0) = -2\text{,}\) and \(s(0)=5\text{,}\) determine the equation for \(s(t)\text{.}\)

2.

Given that \(\int g(x) dx = x^2 + 4x + C\text{,}\) evaluate each indefinite integral below.
  1. \(\displaystyle \int xg(x^2) dx\)
  2. \(\displaystyle \int \frac{g(\ln(x))}{x} dx\)

3.

Evaluate \(\int \tan(x) dx\) using \(u\)-substitution. (Hint: start by writing \(\tan(x)\) as \(\frac{\sin(x)}{\cos(x)}\))

4.

For the town of Calculus, CO, residential carbon emissions have shown certain trends over recent years. Based on data reflecting average emission rates, the town’s environmental agency has modeled emission rates as
\begin{equation*} \frac{dA}{dt} = 4 + \sin(0.263t + 4.7) + \cos(0.526t+9.4)\text{.} \end{equation*}
Here, \(t\) measures time in hours after midnight on a typical weekday, and \(\frac{dA}{dt}\) is the rate of carbon emissions in pounds per hour.
  1. Use technology to sketch a graph of \(\frac{dA}{dt}\) on the interval [0,24] and explain its meaning. Why is this a reasonable model of carbon emissions?
  2. Determine the general solution of this differential equation. What are the units of \(A(t)\text{?}\)
Though technically each individual integral has its own constant, in the end we can add them all together to a single constant. In practice, we typically just add one constant at the very end as we’ve done in this example.
If we recall from the definition of the derivative that \(\frac{du}{dx} \approx \frac{\Delta{u}}{\Delta{x}}\) and use the fact that \(\frac{du}{dx} = g'(x)\text{,}\) then we see that \(g'(x) \approx \frac{\Delta{u}}{\Delta{x}}\text{.}\) Solving for \(\Delta u\text{,}\) \(\Delta u \approx g'(x) \Delta x\text{.}\) It is this last relationship that, when expressed in “differential” notation enables us to write \(du = g'(x) \, dx\) in the change of variable formula.
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