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 23.3.1Separable differential equationsare those which can be written in the form
dy-= f-(x).
dx g(y)
The reason these are called separable is that if you formally cross multiply,
g(y)dy = f (x )dx
and the variables are “separated”. The x variables are on one side and the y variables are on theother.
Proposition 23.3.2If G^{′}
(y)
= g
(y)
and F^{′}
(x)
= f
(x)
, then if the equation, F
(x)
− G
(y)
= cspecifies y as a differentiable function of x, then x → y
= c for various values of c are called integral curvesor solution curves. It makes sense to think of these as giving a solution if, near a point on thelevel curve, one variable is a function of the other.
Example 23.3.4Find the solution to the initial value problem,
2 ′
y y = x,y(0) = 1.
This is a separable equation and in fact, y^{2}dy = xdx so the solution to the differential equation is of the
form
y3
3-
−
2
x2
= C and it only remains to find the constant C. To do this, you use the initial condition.
Letting x = 0, it follows
13
= C and so
y3 x2 1
--− -- = -
3 2 3
The following picture shows how the integral curves follow the tangent field.
PICT
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 23.3.5Find integral curves for the equation
2
y′ =--x----
(1− y2)
Separating variables, you get
( 2)
1− y
dy = x^{2}dx and so the integral curves are of the form
( )
y− y3 − x3 = C
3 3
Here is a picture of a few of these integral curves along with the slope field.
PICT
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
x=[1,2,3]; y= [2,3,4]; x.*y
and then press “enter”. You will get 2,6,12. Of course you would get an error if you wrote x*y. Similarly,
type
[2,4,6,8]./[1,2,3,4]
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
“enter”.
Example 23.3.6What is the equation of a hanging chain?
Consider the following picture of a portion of this chain.
PICT
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
(x)
and T_{0} 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,
T (x)sinθ = l(x)ρg,T (x )cosθ = T0.
Therefore, dividing these yields
sinθ ◜≡ ◞c◟-◝
---- = l(x)ρg∕T0.
cosθ
Now letting y
(x)
denote the y coordinate of the hanging chain corresponding to x,
sin-θ = tan θ = y′(x).
cosθ
Therefore, this yields
′
y (x) = cl(x ).
Now differentiating both sides of the differential equation,
∘ ---------
y′′(x) = cl′(x ) = c 1+ y′(x )2
and so
∘--y′′(x)---= c.
1+ y′(x)2
Let z
(x)
= y^{′}
(x)
so the above differential equation becomes
z′(x )
√-----2 = c.
1 + z
Therefore, ∫
√z′(x)
1+z2
dx = cx + d. Change the variable in the antiderivative letting u = z
(x)
and this
yields
∫ z′(x) ∫ du − 1 −1
√1-+-z2dx = √1-+-u2-= sinh (u)+ C = sinh (z (x))+ C.
Therefore, combining the constants of integration,
−1 ′
sinh (y (x )) = cx + d
and so
′
y (x) = sinh (cx + d).
Therefore,
1
y(x) = -cosh(cx+ d)+ k
c
where d and k are some constants and c = ρg∕T_{0}. Curves of this sort are called catenaries.
Note these curves result from an assumption that the only forces acting on the chain are as
shown.
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
dy = ky
dt
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
( )
1− Ky
which will cause the population growth to decrease as soon as y
exceeds K. 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 23.3.7The equation
dy ( -y)
dt = r 1− K y, r,K > 0
is called the logistic equation.It models population growth. You see that the right side is equal to 0 at thetwo values y = K and y = 0.
This is a separable equation. Thus
dy
(----y)--= rdt
1 − K y
Now you do ∫
to both sides. This requires partial fractions on the left.
1 1 1
(1−--y)y-= K-−-y-+ y-
K
Therefore,
ln(y)− ln(K − y) = rt+ C
if 0 < y < K. If y > K, you get
ln(y)− ln(y− K ) = rt+ C
Therefore, the integral curves are of the form
( y )
ln ------ = rt+ C
K − y
so changing the name of the constant C, it follows that for y < K, the integral curves are described by the
following function.
Cert
y = K Cert +-1, C > 0
In case y > K, these curves are described by
Cert
y = K --rt---, C > 0
Ce − 1
What follows is a picture of the slope field along with some of these integral curves in case r = 1 and
K = 10.
PICT
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 23.3.8Consider the equation
dy
dt
= f
(y)
. Then y_{0}is called an equilibrium point iff
(y0)
= 0. Note that the solution to the initial value problem y^{′} = f
(y)
,y
(t0)
= y_{0}is just y = y_{0}.An equilibriumpoint is stable if whenever y_{1}is close enough to y_{0}, it follows that the solution to theinitial value problem y^{′} = f
(y)
, y
(0)
= y_{1}stays close to y_{0}for all t > 0. It isasymptotically stableif whenever y_{1}is close enough to y_{0}, it follows that for y the solution to the initial value problem,y^{′} = f
(y)
, y
(0)
= y_{1}satisfies lim_{t→∞}y
(t)
= y_{0}. The equilibrium point y_{0}is unstable if there areinitial conditions close to y_{0}but the solution does not stay close to y_{0}. That is, there exists ε > 0
such that for any δ > 0 there is y_{1}with
|y1 − y0|
< δ but the solution to y^{′} = f
(y)
,y
(0)
= y_{1}hasthe property that for some t > 0,
|y(t)− y0|
≥ ε. An equilibrium point y_{0}is semi-stable if it is stablefrom 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)
= 0. If
there is nothing to start with, it can’t grow. However, if you have any other positive number
for y
(0)
, 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
formulae.
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 y_{1},y_{2},y_{3} respectively. Assume you are in the region where y < K. In an actual experiment,
this is where you would be. Let λ ≡ e^{rt}. Then from the above formula for y, you have the
equations
Then divide the second equation by λ and compare with the first. This shows that λ = y_{2}∕y_{1}. Next divide
the top equation by Cλ and the last by Cλ^{3}. This yields
( ) ( )
-1- -1--
K = y1 1+ C λ = y3 1 + Cλ3
Now it becomes possible to solve for C. This yields
( )
C = -y31y3 −-y21y22
y1y32 − y32y3
Then substitute this in to the first equation. This obtains
Then you can solve this for K. After some simplification, it yields
y22y3 − y21y3
-y2−-y-y--= K
2 1 3
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.
Example 23.3.9
dydt
= −r
(1 − yT)
(1− yK-)
y, r > 0,0 < T < K.
This is a separable equation.
(-----)d(y----)--= − rdt
1− yT- 1− yK-y
The partial fractions expansion is
( )
(-----)1(-----)--= ---1-- --K--− -T---- + 1-
1 − yT- 1− yK-y K − T T − y K − y y
Therefore,
-−-1-K ln|T − y|+ --1--T ln|K − y |+ ln|y| = − rt+ C
K − T K − T
Consider the case where r = 1,T = 5,K = 10. Then you get
( || ||)
ln ||(10-−-y)y|| = − t+ C
|(5 − y)2 |
There are cases, depending on where y is. Suppose first that y ∈