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
Example 5.8. Simple First Order Differential Equation.
\(\ds y'=t^2+1\) is a first order linear differential equation; \(\ds F(t,y,y')= y't^21\text{.}\) Show that all solutions to this equation are of the form \(\ds y=t^3/3+t+C\text{.}\)
We first note that \(y = t^3/3 + t + C\) is a solution to the differential equation, since
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:
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 \(\ds\frac{dy}{dx}=2x\text{.}\)
We integrate both sides of the differential equation to find
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:
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:
We had found in Example 5.9 that the solutions to the differential equation were the parabolas
So we use the initial condition to determine the constant \(C\text{:}\)
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 \(\ds f(t)=t^3/3+t+8/3\text{.}\)
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:
Discuss the solutions for this initial value problem.
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
so
that is, the two antiderivatives must be the same except for a constant difference. We can calculate these antiderivatives and rearrange the results:
Here \(\ds A = \pm\, e^C = \pm\, e^{C_0}\) is some nonzero constant. Since we want \(y(0)=40\text{,}\) we substitute and solve for \(A\text{:}\)
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 \(\ds y' = 2(25y)\text{,}\) 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 nonseparable equations as follows.
Definition 5.16. Separable DE.
A first order differential equation is separable if it can be written in the form
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 nonlinear, 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
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
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 nonconstant 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
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
follow these steps to find the nonconstant 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 \(\ds y' = 2t(25y)\text{.}\)
This is almost identical to the previous example. As before, \(y(t)=25\) is a solution. If \(y\not=25\text{,}\)
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
We begin by separating the variables and get
Now integrate both sides to obtain
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
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.
The constant solution is \(y(t)=0\text{;}\) of course this will not be the solution to any interesting initial value problem. For the nonconstant solutions, we proceed much as before:
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
with growth rate constant \(k\) models simple growth and decay of a quantity \(y\) at time \(t\) with solution
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?
Let \(y(t)\) denote the balance in the account at the start of year \(t\text{.}\) Then
for some constant \(k\text{.}\) We can solve this differential equation using separation of variables to obtain
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 \(k\text{:}\)
Therefore,

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 socalled 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 \(r(My)\) 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
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 nonlinear differential equation, notice that the equation is separable
Integrating both sides we obtain
where \(A = \dfrac{My_0}{y_0}\) for \(t_0=0\text{.}\)
Lastly, we solve for \(y\) and get the solution
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
as
Then we can make the following interpretations.

When the quantity \(y\) is small, then the term \(\dfrac{My}{M}\) is close in value to one, and so the differential equation
\begin{equation*} \frac{dy}{dt} = ry(My) \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 \(\dfrac{My}{M}\) 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
with positive growth constant \(r\) and carrying capacity \(M\) models logistic growth of a quantity \(y\) at time \(t\) with solution
where \(A=\dfrac{My_0}{y_0}\) 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?

\(\ds y' = \sin (ty)\)
AnswerNot separable. 
\(\ds y' = e^t e^y\)
AnswerSeparable:\begin{equation*} \frac{dy}{dt} = e^t e^y \implies \frac{dy}{e^y} = e^t \,dt \end{equation*} 
\(\ds yy' = t\)
AnswerSeparable:\begin{equation*} y\frac{dy}{dt} = t \implies y\,dy = t\,dt \end{equation*} 
\(\ds y' = (t^3 t) \arcsin(y)\)
AnswerSeparable:\begin{equation*} \frac{dy}{dt} = (t^3t)\arcsin(y) \implies \frac{dy}{\arcsin(y)} = (t^3t) \,dt \end{equation*} 
\(\ds y' = t^2 \ln y + 4t^3 \ln y\)
AnswerNot separable.
Exercise 5.2.2.
Identify the constant solutions (if any) of the following differential equations.

\(y' =t\sin y\)
AnswerSolution\(y=n\pi\text{,}\) for any integer \(n\text{.}\)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 \(n\text{.}\) 
\(\ds y'=te^y\)
AnswerSolutionnoneAll 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 \(y\text{.}\)

\(\ds y' = 1/(1+t^2)\)
AnswerSolution\(\ds y=\arctan t + C\)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 \(C\text{.}\)

\(y' = \ln t\)
AnswerSolution\(\ds y=t\ln tt+C\)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*} 
\(y' = t/y\)
AnswerSolution\(\ds y=\pm\sqrt{t^2+C}\)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*} 
\(\ds y' = y^2 1\)
AnswerSolution\(\ds y=\pm 1\text{,}\) \(\ds y=(1+Ae^{2t})/(1Ae^{2t})\)This is a first order seperable differential equation:\begin{equation*} \begin{split} \diff{y}{t} \amp= y^21 \\ \int \frac{1}{y^21}\,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^21} \,dy \end{equation*}we use partial fraction decomposition:\begin{equation*} \begin{split}\int \frac{1}{y^21}\,dy \amp= \int \frac{1}{2(y1)} \frac{1}{2(y+1)}\,dy \\ \amp= \frac{1}{2} \left(\lny1  \lny+1\right) + C. \end{split} \end{equation*}Therefore, the general solution (in implicit form) is\begin{equation*} \frac{1}{2} \left(\lny1  \lny+1\right) + C = t, \end{equation*}and \(y= \pm 1 \text{.}\) 
\(\ds y' = t/(y^3  5)\)
AnswerSolution\(\ds y^4/45y=t^2/2+C\)We again notice that the differential equation is separable. Therefore, we compute
\begin{equation*} \begin{split} \diff{y}{t} \amp = \frac{t}{y^35} \\ \int \left(y^35\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 \(C\text{.}\)

\(\ds y'=k(My)\)
AnswerSolution\(\ds y=M+Ae^{kt}\)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(My) \\ \int \frac{1}{My}\,dy \amp= \int k \,dt \\  \ln My \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 \(n\ge 0\)
AnswerSolution\(\ds y={t^{n+1}\over n+1}+1\)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*} 
\(y' = y^{1/3}\text{,}\) \(y(0)=0\)
AnswerSolution\(\ds y=(2t/3)^{3/2}\) and \(y(t)=0\)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 \(y'(0)=3\)
AnswerSolution\(\ds y=2e^{3t/2}\)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\\ \lny \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.
Exercise 5.2.6.
Exercise 5.2.7.
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:
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 halflife of Bismuth is 5 days. Mathematically, this means that
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
Now,
So, after 6 days, we find that there is approximately 261 mg of Bismuth remaining.
Additionally, we wish to find \(t_*\) such that \(y(t_*) = 2\text{:}\)
Thus, after approximately 41 days, there is only 2 mg of Bismuth remaining.
Exercise 5.2.8.
Exercise 5.2.9.
Exercise 5.2.10.
Exercise 5.2.11.
Given the logistic equation \(y' = ky(My)\text{,}\)

Solve the differential equation for \(y\) in terms of \(t\text{.}\)
AnswerSolution\(\ds y={M\over 1+Ae^{Mkt}}\)We separate variables (assuming \(y \neq M,\)):\begin{equation*} \begin{split} \int \frac{dy}{y(My)} \amp= \int k\,dt \\[1ex] \int \left(\frac{1}{y} + \frac{1}{My}\right)\,dy \amp= \int kM \,dt \\[1ex] \ln y  \lnMy \amp= kMt + C \\[1ex] \ln \left\vert \frac{My}{y} \right\vert \amp= kMtC \\[1ex] \left\vert\frac{My}{y}\right\vert \amp= e^{kMtC} \\[1ex] \frac{My}{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{My_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 \(A = \frac{10001}{1} = 999\text{:}\)\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
where time is measured in days.
Solve the differential equation for \(y\) in terms of \(t\text{.}\) Answer
Solution\(y(t) = \frac{64}{1+31e^{50.8416t}} \)We separate variables (assuming \(y \neq 64\)):\begin{equation*} \begin{split} \int \frac{dy}{y(1y/64)} \amp= \int 0.7944\,dt \\[1ex] \int \left(\frac{1}{y} + \frac{1}{64y}\right)\,dy \amp= \int 64(0.7944) \,dt \\[1ex] \ln y  \ln64y \amp= 50.8416t + C \\[1ex] \ln \left\vert \frac{64y}{y} \right\vert \amp= 50.8416tC \\[1ex] \left\vert\frac{64y}{y}\right\vert \amp= e^{50.8416tC} \\[1ex] \frac{64y}{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
SolutionAbout 0.065 days.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*}