Math Insight

Solving pure time differential equations through integration

Math 201, Spring 22
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  1. In order to solve a pure time differential equation, we need to undo the process of taking a derivative. This is much more involved and cannot always be done, but in the case of many simple functions, there are some rules to follow. Before we can learn these, however, we need some notation and terminology. If the derivative of $F(t)$ is $f(t)$, i.e., if $\diff{F}{t} = f$, then we call $F(t)$ an antiderivative of $f(t)$. We've seen that if $F(t)$ is an antiderivative of $f(t)$, then so is $F(t)+C$ for any constant $C$. When we want to talk about all of these functions, we use an indefinite integral: $$\int f(t) \, dt = F(t)+C$$ which is read "the integral of f dt" or "the integral of f with respect to t". We call $f(t)$ the integrand. The $dt$ is a necessary part of the expression and serves to identify the variable with respect to which we are integrating.

    1. Consider the following integral: $$\int 8 t^{3} + 6 t + 5\, dt$$ Recall the rules for taking the derivative of a polynomial: the derivative of $t^{7} + 4 t^{5} + 11 t$ is
      . When we take the derivative of a polynomial, we treat each term separately. For each term, we multiply the coefficient by the
      , and
      . Does the order of these two operations matter?

      If we want to undo taking the derivative of a polynomial, we again treat each term separately, and need to undo both operations in the
      order. We need to
      and then
      .

      Let's look at the first term: $8 t^{3}$. The new exponent is
      , and the new coefficient is
      .

      For the second term, $6 t$, the new exponent is
      , and the new coefficient is
      .

      For the third term, $5=5 t^{0}$, the new exponent is
      , and the new coefficient is
      .

      Then $\displaystyle \int 8 t^{3} + 6 t + 5\, dt=$

    2. Now that we've worked through an example, let's write down a general rule for integrating polynomials. For each term of the form $a t^{n}$, an antiderivative is
      . Are there any values of $n$ for which this doesn't work? If so, what are they? (If not, enter none.)
      Once we've done this for each term, we can find the integral of the polynomial by
      Need help?

    3. Evaluate the following integrals.

      $\displaystyle \int 3 t^{8} + 7 t^{3} + 4 t\, dt=$

      $\displaystyle \int 4 t^{3} + t^{2} + 42\, dt=$

      $\displaystyle \int 5 t^{9} + 7 t^{5} + 4 t^{4} + 3 t^{2} + 10 t\, dt=$

  2. Certain exponential functions also have simple formulas for their antiderivatives.

    1. Let's start by recalling the derivative of the exponential function: the derivative of $e^{t}$ is
      . What is $\displaystyle \int e^{t}\, dt$ then?

      What about the derivative of $e^{c t}$ where $c$ is some fixed non-zero number? The derivative is
      . What is $\displaystyle \int c e^{c t}\, dt$ then?

      For $c \neq 0$, what is the formula for $\displaystyle \int a e^{c t}\, dt$?

      Note: this approach only works if the exponent is no more complicated than a constant times the variable. In other cases, more complicated methods of solving integrals may be required.

    2. Evaluate the following integrals.

      $\displaystyle \int 30 e^{5 t}\, dt=$

      $\displaystyle \int 4 t^{3} + 4 e^{7 t}\, dt=$

      $\displaystyle \int 8 t + e^{3 t} + 3\, dt=$

  3. We will now learn how to evaluate integrals of the form $$\int \frac{a}{b+ct} \, dt$$ where $c \neq 0$.

    1. The formula for integrating a power ($\int t^n dt = \frac{t^{n+1}}{n+1}+C$) does not work if the exponent is $n=$
      . To find an antiderivative for $\displaystyle \int \frac{1}{t}\, dt$, we need to recall what the derivative of $\ln(t)$ is: $\frac{d}{dt} \ln(t) =$

      Based on this, it is natural to expect that $\displaystyle \int \frac{1}{t}\, dt =$
      . There is a subtle problem with this, however. For what values of $t$ is $\frac{1}{t}$ not defined?
      For what values of $t$ is $\ln(t)$ not defined?
      Therefore, $\ln(t)$ is only an antiderivative for positive $t$.

      To get around this problem, consider $\ln (-t)$, which is defined for
      . $\frac{d}{dt} \ln (-t) =$

      So $\ln (t)$ is an antiderivative of $1/t$ for
      and $\ln (-t)$ is an antiderivative of $1/t$ for
      . We can combine these into $\ln{\left (\left\lvert{t}\right\rvert \right )}$ Therefore, $\displaystyle \int \frac{1}{t}\, dt =$
      . Now, we can compute the antiderivative of $1/t$ as long as $t$ is not equal to
      .

      Need help?

    2. Can we use the idea behind $\displaystyle \int \frac{1}{t} \, dt$ to help us evaluate $\displaystyle \int \frac{a}{b+ct} \, dt$? First, let's suppose $b=0$ and we're looking at $\displaystyle \int \frac{a}{ct} \, dt$. This is really just a constant times $\frac{1}{t}$; it is $\frac{a}{ct} =$
      $\frac{1}{t}$. Recall that when we took the derivative of a constant times a function $f$, we just multiplied the derivative of $f$ by the constant. Therefore $\displaystyle \int \frac{a}{ct} \, dt =$

      By the same rationale, we can factor out the $a$ and write $\displaystyle \int \frac{a}{b+ct} \, dt =$
      $\times \displaystyle \int \frac{1}{b+ct} \, dt$. We just need to determine how to integrate $\frac{1}{b+ct}$.

      The chain rule tells us that $\frac{d}{dt} \ln (f(t)) =$
      .
      So $\frac{d}{dt} \ln (b+ct) =$
      .
      If we divide both sides by $c$, we get $\displaystyle \frac{d}{dt}\left[\frac{1}{c}\ln (b+ct)\right] =$
      .
      Now, we can turn this equation around to see what the antiderivative is (though we still need to add an absolute value like we did for the antiderivative of $1/t$). The result is:
      $\displaystyle \int \frac{1}{b+ct} \, dt =$

      We can multiply by $a$ to get back to the problem we started with:
      $\displaystyle \int \frac{a}{b+ct} \, dt =$
      .

      Need help?

    3. Evaluate the following integrals.

      $\displaystyle \int \frac{3}{2 t + 5}\, dt =$

      $\displaystyle \int 2 t + \frac{4}{- 4 t + 7}\, dt =$

      $\displaystyle \int e^{- t} + \frac{9}{3 t - 2}\, dt =$

  4. Now that we have learned how to evaluate several classes of integrals, let's use this to solve some differential equations.

    1. Consider the differential equation \begin{align*} \frac{d x}{d t} &= 3 t^{2} + 4 t - 3\\ x(0) &= 4 \end{align*} We start by finding the indefinite integral of $3 t^{2} + 4 t - 3$, since that is the family of all functions whose derivatives equal $3 t^{2} + 4 t - 3$.

      $\displaystyle \int 3 t^{2} + 4 t - 3\, dt =$

      To determine which one of these functions $x$ is, we plug in our initial condition. According to the initial conditions, $x(0)$ must be
      . If we plug in $t=0$ into our expression for the antiderivative, we get
      . If $x(t)$ is the solution of the differential equation, then these two expressions must be equal, which means that $C=$
      .

      We conclude that our solution to the differential equation is $x(t) =$
      .

    2. The volume of a cell is increasing at a rate of $\frac{10}{2 t + 5}$ $\mu {\rm m}^3$ per minute. If the volume at $t=0$ is $500$ $\mu {\rm m}^3$, find a function describing the volume of the cell and determine the volume of the cell after $10$ minutes.

      Let's solve this problem step by step. First, letting $v(t)$ denote the volume after $t$ minutes, the differential equation describing the volume is \begin{align*} \frac{d v}{d t} &= \frac{10}{2 t + 5} \\ v(0) &= 500 \end{align*}

      We solve this by first finding the indefinite integral
      $\displaystyle \int \frac{10}{2 t + 5}\, dt =$

      Since the initial condition is when $t=0$ and time is moving forward, we know that $t \ge 0$. With this restriction on $t$, the expression $2t+5$ is
      , so we can ignore the absolute values and just consider
      $v(t)=$
      .

      We can now plug in our initial condition to find that $C=$
      . (Use exact values rather than rounding, so $C$ will include a logarithm.)

      Then the function describing the cell volume is $v(t) =$
      , and the population after $10$ minutes will be $v(10) = $
      $\approx$
      $\mu {\rm m}^3$. (Enter the exact answer in the first blank, and a number rounded to at least $4$ significant digits in the second.)

      Need help?

    3. A population of bacteria is growing at a rate of $2 e^{\frac{t}{20}}$ bacteria per hour. If the population after two hours is $300$, what will the population be after twelve hours?

      Letting $p(t)$ denote the population after $t$ hours, write down the differential equation describing the situation:
      $\frac{d p}{d t}=$
      , with $p($
      $)=$
      .

      To find the function describing the bacteria population, we start with the indefinite integral:
      $\displaystyle \int 2 e^{\frac{t}{20}}\, dt =$

      Next, plug in the value of $t$ from the initial condition (i.e., $t=$
      ) into this expression to get
      . Since this value must be equal to the number
      , we solve for $C$ to determine that $C=$
      . (Use exact values rather than rounding.)

      Then the function describing the bacteria population is $p(t) =$
      , and the population after twelve hours will be $p(12) = $
      $\approx$
      bacteria (enter the exact answer in the first blank, and then the nearest whole number in the second).

      Need help?

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Math 201, Spring 22