23.3 Separable Differential Equations, Stability
Separable differential equations also occur quite often in applications and they are fairly easy to deal with.
This section gives a discussion of these equations.
Definition 23.3.1 Separable differential equations are those which can be written in the form
The reason these are called separable is that if you formally cross multiply,
and the variables are “separated”. The x variables are on one side and the y variables are on the
Proposition 23.3.2 If G′
, then if the equation, F
specifies y as a differentiable function of x, then x → y
solves the separable differential
Proof: Differentiate both sides of F
with respect to x.
Using the chain
Therefore, since F′
which is equivalent to
Definition 23.3.3 The curves F
c for various values of c are called integral curves
or solution curves. It makes sense to think of these as giving a solution if, near a point on the
level curve, one variable is a function of the other.
Example 23.3.4 Find the solution to the initial value problem,
This is a separable equation and in fact, y2dy = xdx so the solution to the differential equation is of the
and it only remains to find the constant C.
To do this, you use the initial condition.
The following picture shows how the integral curves follow the tangent field.
Sometimes, you can’t expect to solve for one of the variables in terms of the other. In other
words, the integral curve might not be a function of one variable. Here is a nice example from
Example 23.3.5 Find integral curves for the equation
Separating variables, you get
and so the integral curves are of the form
Here is a picture of a few of these integral curves along with the slope field.
I used MATLAB to graph the above. One thing might be helpful to mention about MATLAB. It is very
good at manipulating matrices and vectors and there is distinctive notation used to accomplish this. For
example say you type
and then press “enter”. You will get 2,6,12. Of course you would get an error if you wrote x*y. Similarly,
and press “enter”. This yields 2,2,2,2. Of course [2,4,6,8]/[1,2,3,4] doesn’t make any sense.
You can get graphs of some integral curves in MATLAB by typing the following and then “enter”. You
don’t have to type it on two lines, but if you want to do so, to get to a new line, you press “shift” and
Example 23.3.6 What is the equation of a hanging chain?
Consider the following picture of a portion of this chain.
In this picture, ρ denotes the density of the chain which is assumed to be constant and g is the
acceleration due to gravity. T
represent the magnitude of the tension in the chain at t
and at 0
respectively, as shown. Let the bottom of the chain be at the origin as shown. If this chain does not move,
then all these forces acting on it must balance. In particular,
Therefore, dividing these yields
Now letting y
coordinate of the hanging chain corresponding to x
Therefore, this yields
Now differentiating both sides of the differential equation,
so the above differential equation becomes
Change the variable in the antiderivative letting u
Therefore, combining the constants of integration,
where d and k are some constants and c = ρg∕T0. Curves of this sort are called catenaries.
Note these curves result from an assumption that the only forces acting on the chain are as
The next example has to do with population models. It was mentioned earlier. The idea is that if there
were infinite resources, population growth would satisfy the differential equation
where k is a constant. However, resources are not infinite and so k should be modified to be consistent with
this. Instead of k, one writes r
which will cause the population growth to decrease as soon as
. Of course the problem with this is that we are not sure whether K
itself is dependent on other
factors not included in the model.
Example 23.3.7 The equation
is called the logistic equation. It models population growth. You see that the right side is equal to 0 at the
two values y = K and y = 0.
This is a separable equation. Thus
Now you do ∫
to both sides. This requires partial fractions on the left.
if 0 < y < K. If y > K, you get
Therefore, the integral curves are of the form
so changing the name of the constant C, it follows that for y < K, the integral curves are described by the
In case y > K, these curves are described by
What follows is a picture of the slope field along with some of these integral curves in case r = 1 and
K = 10.
The bottom axis is the t axis. Note how all the integral curves in the picture approach K as t increases.
This is why K is called a stable equilibrium point.
Definition 23.3.8 Consider the equation
. Then y0 is called an equilibrium point if
. Note that the solution to the initial value problem y′
y0 is just y
An equilibrium point is stable if whenever y1 is close enough to y0, it follows that the solution to the
initial value problem y′
y1 stays close to y0 for all t >
0. It is asymptotically stable
if whenever y1 is close enough to y0, it follows that for y the solution to the initial value problem,
y0. The equilibrium point y0 is unstable if there are
initial conditions close to y0 but the solution does not stay close to y0. That is, there exists ε >
such that for any δ >
0 there is y1 with
< δ but the solution to y′
the property that for some t >
≥ ε. An equilibrium point y0 is semi-stable if it is stable
from one side and unstable from the other.
Now observe that y = 0 is the solution which results if you begin with the initial condition y
= 0. If
there is nothing to start with, it can’t grow. However, if you have any other positive number
then you see that the solution curve approaches the stable point K
. You can see
this, not just by looking at the picture but also by taking the limit as t →∞
in the above
One of the interesting things about this equation is that it is possible to determine K the maximum
capacity, by taking measurements at three equally spaced times. Suppose you do so at times t,2t,3t and
obtain y1,y2,y3 respectively. Assume you are in the region where y < K. In an actual experiment,
this is where you would be. Let λ ≡ ert. Then from the above formula for y, you have the
Then divide the second equation by λ and compare with the first. This shows that λ = y2∕y1. Next divide
the top equation by Cλ and the last by Cλ3. This yields
Now it becomes possible to solve for C. This yields
Then substitute this in to the first equation. This obtains
Then you can solve this for K. After some simplification, it yields
Note how the equilibrium point K was stable in the above example. There were only two equilibrium
points, K and 0. The equilibrium point 0 was unstable because if the integral curve started near 0 but
slightly positive, it tended to increase to K. Here is another harder example. In this example, there are
three equilibrium points.
y, r >
0 < T < K.
This is a separable equation.
The partial fractions expansion is
Consider the case where r = 1,T = 5,K = 10. Then you get
There are cases, depending on where y is. Suppose first that y ∈
Then you get for a different
You could solve this for y if you like and get