Math Insight

Solving pure time differential equations through integration

Math 201, Spring 2015
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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 $f(t)$, 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
    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 $e^{c t}$ (assume $c \neq 0$)?
      . 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 polynomials does not work if the exponent of $t$ is
      . To find an antiderivative for $\displaystyle \int \frac{1}{t}\, dt$, we need to recall what the derivative of $\log t$ is: $\frac{d}{dt} \log t =$

      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 $\log t$ not defined?
      Therefore, $\log t$ is only an antiderivative for positive $t$.

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

      So $\log t$ is an antiderivative for
      and $\log (-t)$ is an antiderivative for
      . We can combine these into $\log{\left (\left\lvert{t}\right\rvert \right )}$ Therefore, $\displaystyle \int \frac{1}{t}\, dt =$
      .

    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 $\frac{1}{t}$: $\frac{a}{ct} =$
      $\frac{1}{t}$. 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, if we can integrate $\frac{1}{b+ct}$, then $\displaystyle \int \frac{a}{b+ct} \, dt$ will just be $\displaystyle \int \frac{1}{b+ct} \, dt$ multiplied by
      .

      The chain rule tells us that $\frac{d}{dt} \log (f(t)) =$
      .
      So $\frac{d}{dt} \log (b+ct) =$

      and $\displaystyle \int \frac{1}{b+ct} \, dt =$

      To summarize, $\displaystyle \int \frac{a}{b+ct} \, dt =$

    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 (don't simplify, and leave terms in the same order as they were in the integral, with the constant added at the end):

      $=$

      Solving for $C$ yields $C=$
      .

      Then $x(t) =$
      .

    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.

      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
      $\displaystyle \int \frac{10}{2 t + 5}\, dt =$

      When $t=0$, $2t+5$ is
      , so we can ignore the absolute values and just consider
      $v(t)=$
      .

      We can now plug in our initial condition to find $C$ (note: unless otherwise specified, use exact values rather than rounding):

      $=$

      So $C=$
      .

      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 then round to $4$ significant figures in the second).

    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 initial condition:

      $=$

      Solving for $C$ gives us $C=$
      .

      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).