First Order Differential Equations

Section 5.2 First Order Differential Equations

In many fields such as physics, biology or business, a relationship is often known or assumed between some unknown quantity and its rate of change, which does not involve any higher derivatives. It is therefore of interest to study first order differential equations in particular.

Definition 5.7. First Order DE.

A first order differential equation is an equation of the form $F(t, y, y')=0\text{.}$ A solution of a first order differential equation is a function $f(t)$ that makes $\ds F(t,f(t),f'(t))=0$ for every value of $t\text{.}$

Here, $F$ is a function of three variables which we label $t\text{,}$ $y\text{,}$ and $y'\text{.}$ It is understood that $y'$ will explicitly appear in the equation although $t$ and $y$ need not. The variable $y$ itself is dependent on $t\text{,}$ hence it is understood that $y'$ must be the derivative of $y$ with respect to $t\text{.}$ Since only the first derivative of $y$ appears, but no higher order derivative, this is a first order differential equation.

Throughout the notes, we use the independent variable $t$ as many applications are based on the independent variable representing time. If no meaning is attributed to the independent variable, we may want to write a first order differential equation in the usual manner as

\begin{equation*} F(x,y,y')=0\text{.} \end{equation*}

Example 5.8. Simple First Order Differential Equation.

$\ds y'=t^2+1$ is a first order linear differential equation; $\ds F(t,y,y')= y'-t^2-1\text{.}$ Show that all solutions to this equation are of the form $\ds y=t^3/3+t+C\text{.}$

Solution

We first note that $y = t^3/3 + t + C$ is a solution to the differential equation, since

\begin{equation*} \frac{d}{dt} \left(\frac{t^3}{3} + t + C\right) = t^2 + 1\text{,} \end{equation*}

for all $C \in \R\text{.}$

We additionally need to show that there are no other solutions. To do so, we integrate the differential equation:

\begin{equation*} y(t) = \int \left(t^2 + 1\right) \, dt = \frac{t^3}{3} + t + C\text{,} \end{equation*}

for some $C \in R\text{.}$

Thus, all solutions to the differential equation are of the form $\ds y=t^3/3+t+C\text{.}$

Example 5.9. Graphical Solution to First Order Differential Equation.

Sketch various solutions to the differential equation $\ds\frac{dy}{dx}=2x\text{.}$

Solution

We integrate both sides of the differential equation to find

\begin{equation*} y = \int 2x \,dx = x^2 + C \end{equation*}

for any constant $C \in \R\text{.}$ This family of solutions are parabolas which are translated vertically, as shown in the graph below taking $C=-2,0,2\text{.}$

Subsection 5.2.1 Initial Value Problems

Definition 5.10. Initial Conditions.

Initial condition(s) are a set of points that the solution (or its derivatives) must satisfy.

Example 5.11. Initial Conditions.

For a differential equation involving a function $f(t)\text{,}$ initial conditions are of the form:

\begin{equation*} f(t_0)=f_0,\qquad f'(t_0)=f_1, \qquad f''(t_0)=f_2, \qquad \ldots etc\text{.} \end{equation*}

Definition 5.12. Initial Value Problem.

An initial value problem (IVP) is a differential equation along with a set of initial conditions.

Example 5.13. First Order Initial Value Problem.

Solve the initial value problem:

\begin{equation*} \frac{dy}{dx}=2x, y(0)=2\text{.} \end{equation*}

Solution

We had found in Example 5.9 that the solutions to the differential equation were the parabolas

\begin{equation*} y(x) = x^2 + C\text{.} \end{equation*}

So we use the initial condition to determine the constant $C\text{:}$

\begin{equation*} y(0) = (0)^2+ C = 2 \implies C = 2\text{.} \end{equation*}

Therefore, the solution to the initial value problem is $y=x^2 + 2\text{,}$ as shown in the graph below.

Example 5.14. Simple Initial Value Problem.

Verify that the initial value problem $\ds y'=t^2+1\text{,}$ $y(1)=4$ has solution $\ds f(t)=t^3/3+t+8/3\text{.}$

Solution

Observe that $f'(t)=t^2+1$ and $f(1)=1^3/2+1+8/3=4$ as required.

The general first order equation is too general, so we can't describe methods that will work on them all, or even a large portion of them. We can make progress with specific kinds of first order differential equations. For example, much can be said about equations of the form $\ds y' = \phi (t, y)$ where $\phi$ is a function of the two variables $t$ and $y\text{.}$ Under reasonable conditions on $\phi\text{,}$ such an equation has a solution and the corresponding initial value problem has a unique solution. However, in general, these equations can be very difficult or impossible to solve explicitly.

Example 5.15. IVP for Newton's Law of Cooling.

Consider this specific example of an initial value problem for Newton's law of cooling:

\begin{equation*} y' = 2(25-y), \ \ y(0)=40\text{.} \end{equation*}

Discuss the solutions for this initial value problem.

Solution

We first note the zero of the equation: If $y(t_0) = 25\text{,}$ the right hand side of the differential equation is zero, and so the constant function $y(t)=25$ is a solution to the differential equation. It is not a solution to the initial value problem, since $y(0)\not=25\text{.}$ (The physical interpretation of this constant solution is that if a liquid is at the same temperature as its surroundings, then the liquid will stay at that temperature.) So long as $y$ is not 25, we can rewrite the differential equation as

\begin{align*} {dy\over dt}{1\over 25-y}\amp = 2\\ {1\over 25-y}\,dy\amp = 2\,dt\text{,} \end{align*}

\begin{equation*} \int {1\over 25-y}\,dy = \int 2\,dt\text{,} \end{equation*}

that is, the two antiderivatives must be the same except for a constant difference. We can calculate these antiderivatives and rearrange the results:

\begin{align*} \int {1\over 25-y}\,dy \amp = \int 2\,dt\\ (-1)\ln|25-y| \amp = 2t+C_0\\ \ln|25-y| \amp = -2t - C_0 = -2t + C\\ |25-y| \amp = e^{-2t+C}=e^{-2t} e^C\\ y-25 \amp = \pm\, e^C e^{-2t}\\ y \amp = 25 \pm e^C e^{-2t} =25+Ae^{-2t}\text{.} \end{align*}

Here $\ds A = \pm\, e^C = \pm\, e^{-C_0}$ is some non-zero constant. Since we want $y(0)=40\text{,}$ we substitute and solve for $A\text{:}$

\begin{equation*} 40 = 25+Ae^0 \implies A = 15\text{.} \end{equation*}

Therefore, $\ds y=25+15 e^{-2t}$ is a solution to the initial value problem. Note that $y$ is never 25, so this makes sense for all values of $t\text{.}$ However, if we allow $A=0$ we get the solution $y=25$ to the differential equation, which would be the solution to the initial value problem if we were to require $y(0)=25\text{.}$ Thus, $\ds y=25+Ae^{-2t}$ describes all solutions to the differential equation $\ds y' = 2(25-y)\text{,}$ and all solutions to the associated initial value problems.

Subsection 5.2.2 Separable Equations

Why could we solve Example 5.15 from the previous section? Our solution depended on rewriting the equation so that all instances of $y$ were on one side of the equation and all instances of $t$ were on the other. Of course, in this case the only $t$ was originally hidden, since we didn't write $dy/dt$ in the original equation. This is not required, however. This idea of being able to separate the independent and dependent variables in a first order differential equation leads to a classification of first order differential equations into separable and non-separable equations as follows.

Definition 5.16. Separable DE.

A first order differential equation is separable if it can be written in the form

\begin{equation*} \frac{dy}{dt} = f(t) g(y)\text{.} \end{equation*}

Let's come back to all first order differential equations on our list from the previous section and decide which ones are separable or not:

$y' = e^x\sec y$ has order 1, is non-linear, is separable
$y'-e^xy+3 = 0$ has order 1, is linear, is not separable
$y'-e^xy = 0$ has order 1, is linear, is separable

As in the examples, we can attempt to solve a separable equation by converting to the form

\begin{equation*} \int {1\over g(y)}\,dy=\int f(t)\,dt\text{.} \end{equation*}

This technique is called separation of variables. The simplest (in principle) sort of separable equation is one in which $g(y)=1\text{,}$ in which case we attempt to solve

\begin{equation*} \int 1\,dy=\int f(t)\,dt\text{.} \end{equation*}

We can do this if we can find an antiderivative of $f(t)\text{.}$

As we have seen so far, a differential equation typically has an infinite number of solutions. Such a solution is called a general solution. A corresponding initial value problem will give rise to just one solution. Such a solution in which there are no unknown constants remaining is called a specific solution.

The general approach to separable equations is as follows: Suppose we wish to solve $y' = f(t) g(y)$ where $f$ and $g$ are continuous functions. If $g(a)=0$ for some $a$ then $y(t)=a$ is a constant solution of the equation, since in this case $y' = 0 = f(t)g(a)\text{.}$ For example, $y' =y^2 -1$ has constant solutions $y(t)=1$ and $y(t)=-1\text{.}$

To find the non-constant solutions, we note that the function $1/g(y)$ is continuous where $g\not=0\text{,}$ so $1/g$ has an antiderivative $G\text{.}$ Let $F$ be an antiderivative of $f\text{.}$ Now we write

\begin{equation*} G(y) = \int {1\over g(y)}\,dy = \int f(t)\,dt=F(t)+C\text{,} \end{equation*}

so $G(y)=F(t)+C\text{.}$ Now we solve this equation for $y\text{.}$

Of course, there are a few places this ideal description could go wrong: Finding the antiderivatives $G$ and $F\text{,}$ and solving the final equation for $y\text{.}$ The upshot is that the solutions to the original differential equation are the constant solutions, if any, and all functions $y$ that satisfy $G(y)=F(t)+C\text{.}$

Guideline for Separation of Variables.

Given the differential equation

\begin{equation*} \frac{dy}{dt} = f(t)g(y)\text{,} \end{equation*}

follow these steps to find the non-constant solutions.

Separate the variables:

\begin{equation*} \frac{dy}{g(y)} = f(t)\,dt \end{equation*}
Apply the integration operator:

\begin{equation*} \int \frac{dy}{g(y)} = \int f(t)\,dt \end{equation*}
If an antiderivative exists for $f$ and for $1/g\text{,}$ and we can solve for $y\text{,}$ then

\begin{equation*} G(y) = F(t) + C \end{equation*}

for some constant $C\text{.}$

Example 5.17. Solving a Separable Differential Equation I.

Solve the differential equation $\ds y' = 2t(25-y)\text{.}$

Solution

This is almost identical to the previous example. As before, $y(t)=25$ is a solution. If $y\not=25\text{,}$

\begin{align*} \int {1\over 25-y}\,dy \amp = \int 2t\,dt\\ (-1)\ln|25-y| \amp = t^2+C_0\\ \ln|25-y| \amp = -t^2 - C_0 = -t^2 + C\\ |25-y| \amp = e^{-t^2+C}=e^{-t^2} e^C\\ y-25 \amp = \pm\, e^C e^{-t^2}\\ y \amp = 25 \pm e^C e^{-t^2} =25+Ae^{-t^2}\text{.} \end{align*}

As before, all solutions are represented by $\ds y=25+Ae^{-t^2}\text{,}$ allowing $A$ to be zero.

Example 5.18. Solving a Seperable Differential Equation II.

Find the solutions to the differential equation

\begin{equation*} \sec(t) \frac{dy}{dt} - e^{y+\sin(t)} = 0\text{.} \end{equation*}

Solution

We begin by separating the variables and get

\begin{equation*} \begin{split} \sec(t) \frac{dy}{dt} \amp = e^{y+\sin(t)} \\ \sec(t) \frac{dy}{dt} \amp = e^ye^{\sin(t)} \\ e^{-y}\,dy \amp = \frac{e^{\sin(t)}}{\sec(t)}\,dt = \cos(t)e^{\sin(t)}\,dt \end{split} \end{equation*}

Now integrate both sides to obtain

\begin{equation*} \begin{split} \int e^{-y}\,dy \amp = \int \cos(t)e^{\sin(t)}\,dt \\ -e^{-y} \amp = e^{\sin(t)} + C \\ y \amp = - \ln\left(D - e^{\sin(t)}\right) \end{split} \end{equation*}

For convenience, we left out the absolute value in the argument of the logarithm. As in the previous examples, care must be taken to ensure that the argument of the logarithm is positive for a given value of $D\text{.}$

Therefore, the solutions to the differential equation are given by

\begin{equation*} y = - \ln\left(D - e^{\sin(t)}\right)\text{,} \end{equation*}

for some constant $D\text{.}$

Subsection 5.2.3 Simple Growth and Decay Model

Example 5.19. Rate of Change Proportional to Size.

Find the solutions to the differential equation $y'=ky\text{,}$ which models a quantity $y$ that grows or decays proportionally to its size depending on whether $k$ is positive or negative.

Solution

The constant solution is $y(t)=0\text{;}$ of course this will not be the solution to any interesting initial value problem. For the non-constant solutions, we proceed much as before:

\begin{align*} \int {1\over y}\,dy\amp = \int k\,dt\\ \ln|y| \amp = kt+C\\ |y| \amp = e^{kt} e^C\\ y \amp = \pm \,e^C e^{kt}\\ y\amp = Ae^{kt}\text{.} \end{align*}

Again, if we allow $A=0$ this includes the constant solution, and we can simply say that $\ds y=Ae^{kt}$ is the general solution. With an initial value we can easily solve for $A$ to get the solution of the initial value problem. In particular, if the initial value is given for time $t=0\text{,}$ $y(0)=y_0\text{,}$ then $A=y_0$ and the solution is $\ds y= y_0 e^{kt}\text{.}$

The constant $k$ in the above differential equation is referred to as the growth rate constant. Furthermore, this type of differential equation is known as a simple model for growth and decay of some quantity, since it only considers that the growth rate is proportional to the size of the quantity itself without any other factors influencing $y\text{.}$ The graph below shows the typical $J$-shape of such a solution for some $y_0\text{.}$

Simple Growth and Decay Model.

The differential equation

\begin{equation*} \frac{dy}{dt} = ky \end{equation*}

with growth rate constant $k$ models simple growth and decay of a quantity $y$ at time $t$ with solution

\begin{equation*} y = y_0e^{kt}\text{,} \end{equation*}

where $y_0$ is the initial value at time $t=0\text{.}$

When $k>0\text{,}$ this describes certain simple cases of population growth: It says that the change in the population $y$ is proportional to the population. The underlying assumption is that each organism in the current population reproduces at a fixed rate, so the larger the population the more new organisms are produced. While this is too simple to model most real populations, it is useful in some cases over a limited time.

When $k\lt 0\text{,}$ the differential equation describes a quantity that decreases in proportion to the current value. This can be used to model radioactive decay.

Interactive Demonstration. Use the sliders below to investigate the differential equation $\frac{dy}{dt} = f(t,y) $ where

Note: The simple growth and decay model is unrestricted because the quantity $y$ grows without bound as $t \to \infty$ if $k > 0\text{.}$ In the decay case, the solution only becomes unbounded if time $t$ is allowed to approach $-\infty\text{.}$

Example 5.20. Simple Growth Model.

Suppose $5,000 is deposited into an account which earns continuously compounded interest. Under these conditions, the balance in the account grows at a rate proportional to the current balance. Suppose that after 4 years the account is worth $7,000.

How much is the account worth after 5 years?
How many years does it take for the balance to double?

Solution

Let $y(t)$ denote the balance in the account at the start of year $t\text{.}$ Then

\begin{equation*} \frac{dy}{dt} = ky(t)\text{,} \end{equation*}

for some constant $k\text{.}$ We can solve this differential equation using separation of variables to obtain

\begin{equation*} y(t)=y_0e^{kt} = 5000e^{kt}\text{,} \end{equation*}

where we used the fact that $y(0)=5,000\text{.}$ We know that $y(4)=7000\text{,}$ which we now use to solve for $k\text{:}$

\begin{equation*} \begin{split} 7000 \amp = 5000e^{4k} \\ e^{4k} \amp =\frac{7}{5}\\ k \amp = \frac{\ln\left(\frac{7}{5}\right)}{4} \end{split} \end{equation*}

Therefore,

\begin{equation*} y(t) = 5000e^{\frac{\ln\left(7/5\right)}{4}t}\text{.} \end{equation*}

After 5 years, the balance in the account is

\begin{equation*} y(5)=5000e^{\frac{\ln\left(7/5\right)}{4} \cdot 5} \approx \$ 7614.30\text{.} \end{equation*}
We wish to find $t$ so that $y(t)=10000\text{.}$

\begin{equation*} \begin{split} 10000 \amp = 5000e^{\frac{\ln\left(7/5\right)}{4}t} \\ e^{\frac{\ln\left(7/5\right)}{4}t} \amp = 2 \\ t \amp = \frac{\ln(16)}{\ln\left(\frac{7}{5}\right)} \approx 8.24 \end{split} \end{equation*}

Thus, it takes just over 8 years for the balance in the account to double in value.

Subsection 5.2.4 Logistic Growth Model

The simple growth model is unrealistic because a quantity that represents something from real life, say, population, does not grow unrestricted. Typically, food resources, competition, predators, or diseases, to name but a few factors, influence the growth of the population, and how much of the population can be sustained in such an environment.

A more realistic model is the so-called logistic growth model, which mimics that as a quantity is growing other factors will influence the growth and slow it down until a certain maximum size is being approached. For example, if a population is growing, then food may become scarce or diseases may break out among the population, and the population growth slows down until a certain sustainable size is reached. Replacing $k$ in the simple model with $r(M-y)$ achieves that as the quantity $y$ increases the growth rate decreases, and furthermore, that the maximum population size that is sustained is $M\text{.}$ This maximum is referred to as the carrying capacity. So the simple model becomes

\begin{equation*} \frac{dy}{dt} = ry(M-y) \end{equation*}

for some positive $r$ and $M\text{.}$

In other words, the rate of growth of the quantity $y$ is proportional to both itself and the remaining carrying capacity that the quantity can still grow to. As usual, $r$ is the growth rate constant.

To solve this first order non-linear differential equation, notice that the equation is separable

\begin{equation*} \frac{dy}{y(M-y)} = r\,dt\text{.} \end{equation*}

Integrating both sides we obtain

\begin{equation*} \begin{split} \int \frac{dy}{y(M-y)} \amp = \int r\,dt \\[1ex] \int \left(\frac{1}{y} + \frac{1}{M-y}\right)\,dy \amp = \int rM \,dt \\[1ex] \ln |y| - \ln|M-y| \amp = rMt + C \\[1ex] \ln \left\vert \frac{M-y}{y} \right\vert \amp = -rMt-C \\[1ex] \left\vert\frac{M-y}{y}\right\vert \amp = e^{-rMt-C} \\[1ex] \frac{M-y}{y} \amp = Ae^{-rMt}, \end{split} \end{equation*}

where $A = \dfrac{M-y_0}{y_0}$ for $t_0=0\text{.}$

Lastly, we solve for $y$ and get the solution

\begin{equation*} y=\frac{M}{1+Ae^{-rMt}} \text{ with } A=\frac{M-y_0}{y_0} \end{equation*}

for some constant $y_0$ at time $t_0=0\text{.}$

The graph below shows the typical $S$-shape of such a solution for some $y_0$ between $0$ and $M\text{.}$ For $y_0 \ge M\text{,}$ the solution decays exponentially to $M\text{.}$ And for $y_0 = M\text{,}$ the solution remains constant.

Note: Let us rewrite

\begin{equation*} \frac{dy}{dt} = ry(M-y) \end{equation*}

\begin{equation*} \frac{dy}{dt} = rM\left(\frac{M-y}{M}\right)y\text{.} \end{equation*}

Then we can make the following interpretations.

When the quantity $y$ is small, then the term $\dfrac{M-y}{M}$ is close in value to one, and so the differential equation

\begin{equation*} \frac{dy}{dt} = ry(M-y) \approx rMy\text{.} \end{equation*}

In other words, the growth is exponential.
However, when the quantity $y$ is near that of the carrying capacity $M\text{,}$ then the term $\dfrac{M-y}{M}$ is close in value to zero. Hence, the less carrying capacity that remains the more the growth rate is slowed down.

Logistic Growth Model.

The differential equation

\begin{equation*} \frac{dy}{dt} = ry(M-y) \end{equation*}

with positive growth constant $r$ and carrying capacity $M$ models logistic growth of a quantity $y$ at time $t$ with solution

\begin{equation*} y=\frac{M}{1+Ae^{-rMt}}\text{,} \end{equation*}

where $A=\dfrac{M-y_0}{y_0}$ at time $t_0=0\text{.}$

Interactive Demonstration. Use the sliders below to investigate the differential equation $\frac{dy}{dt} = f(t,y) $ where

Exercises for Section 5.2.

Exercise 5.2.1.

Which of the following equations are separable?

$\ds y' = \sin (ty)$
Answer
Not separable.
$\ds y' = e^t e^y$
Answer
Separable:
\begin{equation*} \frac{dy}{dt} = e^t e^y \implies \frac{dy}{e^y} = e^t \,dt \end{equation*}
$\ds yy' = t$
Answer
Separable:
\begin{equation*} y\frac{dy}{dt} = t \implies y\,dy = t\,dt \end{equation*}
$\ds y' = (t^3 -t) \arcsin(y)$
Answer
Separable:
\begin{equation*} \frac{dy}{dt} = (t^3-t)\arcsin(y) \implies \frac{dy}{\arcsin(y)} = (t^3-t) \,dt \end{equation*}
$\ds y' = t^2 \ln y + 4t^3 \ln y$
Answer
Not separable.

Exercise 5.2.2.

Identify the constant solutions (if any) of the following differential equations.

$y' =t\sin y$
Answer
$y=n\pi\text{,}$ for any integer $n\text{.}$
Solution
All constant solutions of the DE will satisfy $y'=0$ for all $t \geq 0\text{.}$ Therefore, we set
\begin{equation*} y' = t\sin(y) = 0 \implies \arcsin(0) = y, \end{equation*}
since $t\neq 0\text{.}$ Therefore, the constant solutions are $y=n\pi\text{,}$ for any integer $n\text{.}$
$\ds y'=te^y$
Answer
none
Solution
All constant solutions of the DE will satisfy $y'=0$ for all $t \geq 0\text{.}$ Therefore, we set
\begin{equation*} y' = te^y = 0 \implies e^y = 0, \end{equation*}
which has no solutions. Therefore, the DE has no constant solutions.

Exercise 5.2.3.

Solve the following differential equations. You may leave your solution in implicit form: that is, you may stop once you have done the integration, without solving for $y\text{.}$

$\ds y' = 1/(1+t^2)$
Answer
$\ds y=\arctan t + C$
Solution

This is a first order separable differential equation:

\begin{equation*} \begin{split} \diff{y}{t} \amp = \frac{1}{1+t^2} \\ \int dy \amp = \int \frac{dt}{1+t^2} \\ y \amp = \tan^{-1}(t) + C \end{split} \end{equation*}

Therefore, the general solution is

\begin{equation*} y(t) = \tan^{-1}(t) + C \end{equation*}

for some constant $C\text{.}$
$y' = \ln t$
Answer
$\ds y=t\ln t-t+C$
Solution
This is a first order seperable differential equation:
\begin{equation*} \begin{split} \diff{y}{t} \amp= \ln t \\ \int \,dy \amp= \int \ln t\,dt \\ y(t) \amp= t\ln t - t + C. \end{split} \end{equation*}
$y' = t/y$
Answer
$\ds y=\pm\sqrt{t^2+C}$
Solution
This is a first order seperable differential equation:
\begin{equation*} \begin{split} \diff{y}{t} \amp= \frac{t}{y} \\ \int y\,dy \amp= \int t\,dt \\ \frac{y^2}{2} \amp= \frac{t^2}{2} + C, \end{split} \end{equation*}
or,
\begin{equation*} y=\pm\sqrt{t^2+C}\text{.} \end{equation*}
$\ds y' = y^2 -1$
Answer
$\ds y=\pm 1\text{,}$ $\ds y=(1+Ae^{2t})/(1-Ae^{2t})$
Solution
This is a first order seperable differential equation:
\begin{equation*} \begin{split} \diff{y}{t} \amp= y^2-1 \\ \int \frac{1}{y^2-1}\,dy \amp= \int \,dt \end{split} \end{equation*}
However, this is not valid if $y=\pm 1\text{.}$ In this case, we have $y' = 0\text{,}$ and so these are constant solutions of the DE. Now suppose $y\neq \pm 1\text{.}$ To solve the integral
\begin{equation*} \int \frac{1}{y^2-1} \,dy \end{equation*}
we use partial fraction decomposition:
\begin{equation*} \begin{split}\int \frac{1}{y^2-1}\,dy \amp= \int \frac{1}{2(y-1)}- \frac{1}{2(y+1)}\,dy \\ \amp= \frac{1}{2} \left(\ln|y-1| - \ln|y+1|\right) + C. \end{split} \end{equation*}
Therefore, the general solution (in implicit form) is
\begin{equation*} \frac{1}{2} \left(\ln|y-1| - \ln|y+1|\right) + C = t, \end{equation*}
and $y= \pm 1 \text{.}$
$\ds y' = t/(y^3 - 5)$
Answer
$\ds y^4/4-5y=t^2/2+C$
Solution

We again notice that the differential equation is separable. Therefore, we compute

\begin{equation*} \begin{split} \diff{y}{t} \amp = \frac{t}{y^3-5} \\ \int \left(y^3-5\right)\,dy \amp = \int t\,dt \\ \frac{1}{4}y^4 - 5y \amp = \frac{1}{2}t^2 + C \end{split} \end{equation*}

Hence, the general solution in implicit form is

\begin{equation*} \frac{1}{4}y(t)^4 - 5y(t) = \frac{1}{2}t^2 + C \end{equation*}

for some constant $C\text{.}$
$\ds y'=k(M-y)$
Answer
$\ds y=M+Ae^{-kt}$
Solution
This is a first order seperable differential equation. First, notice that $y=M$ is a constant solution. Now suppose that $y\neq M$ and separate:
\begin{equation*} \begin{split} \diff{y}{t} \amp= k(M-y) \\ \int \frac{1}{M-y}\,dy \amp= \int k \,dt \\ - \ln |M-y| \amp= kt + C \end{split} \end{equation*}
This is the general solution in implicit form. Note that we can further write $y$ explicitly as
\begin{equation*} y(t) = M+Ae^{-kt}, \end{equation*}
for some constant $A\text{.}$ For $A=0$ we recover the solution $y=M\text{.}$

Exercise 5.2.4.

Solve the following initial value problems.

$y' = t^n\text{,}$ $y(0)=1$ and $n\ge 0$
Answer
$\ds y={t^{n+1}\over n+1}+1$
Solution
We find the general solution by separating variables:
\begin{equation*} \begin{split} \diff{y}{t} \amp= t^n \\ \int \,dy \amp= \int t^n \,dt\\ y(t) \amp= \frac{t^{n+1}}{n+1} + C \end{split} \end{equation*}
Therefore, if $y(0) = 1\text{,}$ we require
\begin{equation*} y(0) = 0 + C = 1 \implies C = 1. \end{equation*}
Therefore, for any $n\geq 0\text{,}$ the solution to this IVP is
\begin{equation*} y(t) = \frac{t^{n+1}}{n+1} + 1. \end{equation*}
$y' = y^{1/3}\text{,}$ $y(0)=0$
Answer
$\ds y=(2t/3)^{3/2}$ and $y(t)=0$
Solution
We note that $y(t) = 0$ is a constant solution to this DE, and further satisfies the initial condition. Therefore, a solution to this IVP is $y(t) = 0\text{.}$ We look for any more solutions by separating variables (assuming $y\neq 0$):
\begin{equation*} \begin{split} \diff{y}{t} \amp= y^{1/3} \\ \int y^{-1/3}\,dy \amp= \int \,dt\\ \frac{3y^{2/3}}{2} \amp= t + C \\ y(t) \amp= \frac{2}{3} t^{3/2} + C. \end{split} \end{equation*}
Since $y(0) = 0\text{,}$ we see that we must have $C=0\text{.}$ Therefore, another solution to the IVP is
\begin{equation*} y(t) = \frac{2}{3} t^{3/2} . \end{equation*}
$y' = ky\text{,}$ $y(0)=2\text{,}$ and $y'(0)=3$
Answer
$\ds y=2e^{3t/2}$
Solution
We find the general solution by separating variables:
\begin{equation*} \begin{split} \diff{y}{t} \amp= ky \\ \int \frac{1}{y} \,dy \amp= \int k \,dt\\ \ln|y| \amp= kt + C \\ y(t) \amp= Ae^{kt} \end{split} \end{equation*}
for some constant $A\text{.}$ Since $y(0) =2\text{,}$ we require
\begin{equation*} y(0) = Ae^{k(0)} = A = 2. \end{equation*}
Now if $y'(0) = 3\text{,}$ we require
\begin{equation*} y'(0) = 2 ke^{k(0)} = 2k = 3 \implies k = \frac{3}{2}. \end{equation*}
Therefore, the solution to the IVP is
\begin{equation*} y(t) = 2e^{3t/2}. \end{equation*}

Exercise 5.2.5.

After 10 minutes in Jean-Luc's room, his tea has cooled to $40^\circ$ Celsius from $100^\circ$ Celsius. The room temperature is $25^\circ$ Celsius. How much longer will it take to cool to $35^\circ\text{?}$ Answer

$\approx 2$ minutes

Solution

By Newton's Law of Cooling, the temperature of Jean-Luc's tea can be described by the DE

\begin{equation*} \diff{y}{t} = k (25-y), \end{equation*}

for some $k\text{,}$ where 25 is the ambient room temperature, such that $y(0) = 100$ and $y(10) = 40$ (with $t$ measured in minutes). We first solve for the general solution of this DE. Suppose that $y\neq 25$ (we note that this is not a solution to the IVP). Then we can separate variables:

\begin{equation*} \begin{split} \diff{y}{t} \amp= k (25-y) \\ \int \frac{1}{25-y} \,dy \amp= \int k\,dt \\ -\ln |25-y| \amp= kt + C\\ y \amp= 25 + Ae^{-kt},\end{split} \end{equation*}

for some constants $A$ and $k\text{.}$ We now use the given data to find these constants:

\begin{equation*} y(0) = 100 \implies 25+ A = 100 \implies A = 75 \end{equation*}

and

\begin{equation*} y(10) = 40 \implies 75e^{-10k} = 40 \implies k = \frac{1}{10} \ln(15/8). \end{equation*}

All together, we find that the temperature of the tea can be described by the function

\begin{equation*} y(t) = 75e^{-\ln(15/8)/10 t}. \end{equation*}

We now solve for $t\text{:}$

\begin{equation*} y(t) = 35 \implies 75e^{-\ln(15/8) t/10} = 35 \implies t = \frac{10\log(15/7)}{\log(15/8)} \approx 12.124. \end{equation*}

Hence, it will take about 2 more minutes for the tea to cool down to 35 degrees.

Exercise 5.2.6.

A radioactive substance obeys the equation $y' =ky$ where $k\lt 0$ and $y$ is the mass of the substance at time $t\text{.}$ Suppose that initially, the mass of the substance is $y(0)=M>0\text{.}$ At what time does half of the mass remain? (This is known as the half life. Note that the half life depends on $k$ but not on $M\text{.}$) Answer

$\ds t=-{\ln 2\over k}$

Solution

We first find the general solution to the DE. Suppose that $y\neq 0$ (since we assume that $M>0\text{,}$ this is not a solution to the IVP). Now separate variables:

\begin{equation*} \begin{split} \diff{y}{t} \amp= ky \\ \int \frac{dy}{y} \amp= \int k\, dt \\ \ln |y| \amp= kt + C \\ y(t) \amp= Ae^{kt} \end{split} \end{equation*}

for some constants $A$ and $k\text{.}$ Now apply the initial data:

\begin{equation*} y(0) = M \implies A = M. \end{equation*}

Therefore,

\begin{equation*} y(t) = Me^{kt} \end{equation*}

is the solution to the IVP. We now want to find the time it takes for the initial mass to be reduced by one half. That is, we want to solve for $t^*$ where $y(t^*) = \frac{M}{2}\text{:}$

\begin{equation*} \begin{split} y(t^*) \amp= Me^{kt^*} = \frac{M}{2}\\ \implies e^{kt^*} \amp= \frac{1}{2}\\ kt^* \amp= \ln(1/2)\\ t^* \amp= \frac{1}{k} \ln(1/2). \end{split} \end{equation*}

Therefore, the half-life is $t^* = -\frac{\ln(2)}{k}\text{.}$ Note that since $k\le 0\text{,}$ we have $t^*\ge 0$ as required.

Exercise 5.2.7.

Bismuth-210 has a half life of five days. If there is initially 600 milligrams, how much is left after 6 days? When will there be only 2 milligrams left? Answer

$\ds 600e^{-6\ln 2/5}\approx 261$ mg; $\ds {5\ln 300\over\ln2}\approx 41$ days

Solution

We model the decay of a radioactive substance by the equation $y'=ky\text{.}$ Therefore, let $y(t)$ be the amount of Bismuth remaining, measured in mg, where $t$ is in days. We first solve the initial value problem:

\begin{equation*} \begin{split} \diff{y}{t} \amp = ky \\ \int \frac{dy}{y} \amp = \int k\, dt \\ \ln |y| \amp = kt + C \\ y(t) \amp = Ae^{kt} \end{split} \end{equation*}

Applying the initial data, $y(0)=600\text{,}$ we see that $y(t)=600e^{kt}\text{.}$ To solve for $k\text{,}$ we use the fact that the half-life of Bismuth is 5 days. Mathematically, this means that

\begin{equation*} \begin{split} y(5) \amp = \frac{1}{2} y(0) \\ e^{5k} \amp = \frac{1}{2} \\ \implies k \amp = \frac{1}{5} \ln\left(\frac{1}{2}\right) = -\frac{\ln 2}{5}. \end{split} \end{equation*}

Notice that $k \lt 0$ and that the quantity of Bismuth is decreasing with time, as desired.

Therefore, the solution to the initial value problem is

\begin{equation*} y(t) = 600e^{-\frac{\ln 2}{5}t}\text{.} \end{equation*}

Now,

\begin{equation*} y(6)=600e^{-\frac{\ln 2}{5}\cdot 6} \approx 261\text{.} \end{equation*}

So, after 6 days, we find that there is approximately 261 mg of Bismuth remaining.

Additionally, we wish to find $t_*$ such that $y(t_*) = 2\text{:}$

\begin{equation*} 2 = 600e^{-t_*\frac{\ln 2}{5}} \implies t_* = \frac{5\ln 300}{\ln 2} \approx 41\text{.} \end{equation*}

Thus, after approximately 41 days, there is only 2 mg of Bismuth remaining.

Exercise 5.2.8.

The half life of carbon-14 is 5730 years. If one starts with 100 milligrams of carbon-14, how much is left after 6000 years? How long do we have to wait before there is less than 2 milligrams? Answer

$\ds 100e^{-200\ln 2/191}\approx 48$ mg; $\ds {5730\ln 50\over\ln2}\approx 32339$ years.

Solution

We model the decay of a radioactive substance by the equation $y'=ky\text{.}$ Therefore, let $y(t)$ be the amount of Bismuth remaining, measured in mg, where $t$ is in years. We first solve the initial value problem:

\begin{equation*} \begin{split} \diff{y}{t} \amp= ky \\ \int \frac{dy}{y} \amp= \int k\, dt \\ \ln |y| \amp= kt + C \\ y(t) \amp= Ae^{kt} \end{split} \end{equation*}

Applying the initial data, $y(0)=100\text{,}$ we see that $y(t)=100e^{kt}\text{.}$ To solve for $k\text{,}$ we use the fact that the half-life of carbon-14 is 5730 years. This means that

\begin{equation*} \begin{split} y(5730) \amp= \frac{1}{2} y(0) \\ e^{5730k} \amp= \frac{1}{2} \\ \implies k \amp= \frac{1}{5730} \ln\left(\frac{1}{2}\right) = -\frac{\ln 2}{5730}. \end{split} \end{equation*}

Notice that $k \le 0$ and that the quantity of carbon-14 is decreasing with time, as desired. \\ Therefore, the solution to the initial value problem is

\begin{equation*} y(t) = 100e^{-\frac{\ln 2}{5730}t}. \end{equation*}

Now,

\begin{equation*} y(6000)=100e^{-\frac{\ln 2}{5730}\cdot 6000} \approx 48.39. \end{equation*}

So, after 6000 years, we find that there is approximately 48.39 mg of carbon-14 remaining (note that this is just under half of the initial amount, as expected). Additionally, we wish to find $t_*$ such that $y(t_*) = 2\text{:}$

\begin{equation*} 2 = 100e^{-t_*\frac{\ln 2}{5730}} \implies t_* = \frac{5730\ln(50)}{\ln(2)} \approx 32,339. \end{equation*}

Therefore, we have to wait more than 32,339 years for the amount carbon-14 to reduce to less than 2 mg.

Exercise 5.2.9.

A certain species of bacteria doubles its population (or its mass) every hour in the lab. The differential equation that models this phenomenon is $y' =ky\text{,}$ where $k>0$ and $y$ is the population of bacteria at time $t\text{.}$ What is $y\text{?}$ Answer

$\ds y=y_0e^{\ln(2)t}$

Solution

We model the growth of bacteria by the equation $y'=ky$ where $k>0\text{.}$ We first solve for the general solution assuming that $y\neq 0\text{:}$

\begin{equation*} \begin{split} \diff{y}{t} \amp= ky \\ \int \frac{dy}{y} \amp= \int k\, dt \\ \ln |y| \amp= kt + C \\ y(t) \amp= Ae^{kt} \end{split} \end{equation*}

Suppose that initially we have $y_0$ bacteria present: $y(0) = y_0 > 0\text{.}$ Then, the number of bacteria at time $t$ can be described by the function $y(t)= y_0 e^{kt}.$ Since $k>0\text{,}$ the amount of bacteria is growing exponetially. Now use the fact that the bacteria doubles in growth every hour:

\begin{equation*} y(1) = 2y_0 \implies e^{k(1)} = 2 \implies k = \ln(2). \end{equation*}

Therefore, the growth of this particular species can be described by the function

\begin{equation*} y(t)= y_0 e^{\ln(2) t}. \end{equation*}

Exercise 5.2.10.

If a certain microbe doubles its population every 4 hours and after 5 hours the total population has mass 500 grams, what was the initial mass? Answer

$\ds 500e^{-5\ln2/4}\approx 210$ g

Solution

We model the growth of bacteria by the equation $y'=ky$ where $k>0\text{,}$ where $t$ is measured in hours and $y$ is measured in grams. We first solve for the general solution assuming that $y\neq 0\text{:}$

\begin{equation*} \begin{split} \diff{y}{t} \amp= ky \\ \int \frac{dy}{y} \amp= \int k\, dt \\ \ln |y| \amp= kt + C \\ y(t) \amp= Ae^{kt}\\ \amp= y_0 e^{kt} \end{split} \end{equation*}

Now use the fact that the bacteria doubles in growth every 4 hours:

\begin{equation*} y(4) = 2y_0 \implies e^{k(4)} = 2 \implies k = \frac{1}{4}\ln(2). \end{equation*}

After 5 hours total, the population has 500 grams:

\begin{equation*} y(5) = 500 = y_0 e^{\frac{5}{4}\ln(2)} \implies y_0 = 500e^{-5\ln2/4} \approx 210. \end{equation*}

Therefore, the initial mass must have been about 210 g.

Exercise 5.2.11.

Given the logistic equation $y' = ky(M-y)\text{,}$

Solve the differential equation for $y$ in terms of $t\text{.}$
Answer
$\ds y={M\over 1+Ae^{-Mkt}}$
Solution
We separate variables (assuming $y \neq M,$):
\begin{equation*} \begin{split} \int \frac{dy}{y(M-y)} \amp= \int k\,dt \\[1ex] \int \left(\frac{1}{y} + \frac{1}{M-y}\right)\,dy \amp= \int kM \,dt \\[1ex] \ln |y| - \ln|M-y| \amp= kMt + C \\[1ex] \ln \left\vert \frac{M-y}{y} \right\vert \amp= -kMt-C \\[1ex] \left\vert\frac{M-y}{y}\right\vert \amp= e^{-kMt-C} \\[1ex] \frac{M-y}{y} \amp= Ae^{-kMt}, \end{split} \end{equation*}
Therefore, we find that
\begin{equation*} y=\frac{M}{1+Ae^{-kMt}} \quad\text{with}\quad A=\frac{M-y_0}{y_0} \end{equation*}
Sketch the graph of the solution to this equation when $M=1000\text{,}$ $k=0.002\text{,}$ $y(0)=1\text{.}$ Answer
When $M=1000\text{,}$ $k=0.002$ and $y(0) =1\text{,}$ we have $A = \frac{1000-1}{1} = 999\text{:}$
\begin{equation*} y(t) = \frac{1000}{1+999e^{-2t}} \end{equation*}
as shown below:

Exercise 5.2.12.

The biologist G. F. Gause studied the growth of the protozoan Paramecium in the early 1930s. Through his data, he figured out that the relative growth rate is 0.7944 when $y(0)=2\text{,}$ and the carrying capacity is 64. This leads to the logistic model

\begin{equation*} \frac{dy}{dt}=0.7944y\left(1-\frac{y}{64}\right), y(0)=2\text{,} \end{equation*}

where time is measured in days.

Solve the differential equation for $y$ in terms of $t\text{.}$ Answer
$y(t) = \frac{64}{1+31e^{-50.8416t}} $
Solution
We separate variables (assuming $y \neq 64$):
\begin{equation*} \begin{split} \int \frac{dy}{y(1-y/64)} \amp= \int 0.7944\,dt \\[1ex] \int \left(\frac{1}{y} + \frac{1}{64-y}\right)\,dy \amp= \int 64(0.7944) \,dt \\[1ex] \ln |y| - \ln|64-y| \amp= 50.8416t + C \\[1ex] \ln \left\vert \frac{64-y}{y} \right\vert \amp= -50.8416t-C \\[1ex] \left\vert\frac{64-y}{y}\right\vert \amp= e^{-50.8416t-C} \\[1ex] \frac{64-y}{y} \amp= Ae^{-50.8416t}, \end{split} \end{equation*}
Therefore, we find that
\begin{equation*} y=\frac{64}{1+Ae^{-50.8416t}} \quad\text{with}\quad A=\frac{62}{2} = 31. \end{equation*}
How long will it take for the protozoa to reach 30? Answer
About 0.065 days.
Solution
We solve for $t\text{:}$
\begin{equation*} y(t) = 30 \implies t \approx 0.065 \text{ days}. \end{equation*}
Sketch the graph of the solution to this equation. Answer
We plot the solution
\begin{equation*} y(t) = \frac{64}{1+31e^{-50.8416t}}. \end{equation*}