First Order Differential Equations
Ordinary Differential Equations
Dr. Marco A Roque Sol
12/01/2015
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Modeling
Linear and Nonlinear Equations
Autonomous Equations
Exact Equations
First Order Differential Equations
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Modeling
Linear and Nonlinear Equations
Autonomous Equations
Exact Equations
Modeling
Example 20
Newton’s Law of Cooling
Newton’s Law of Cooling states that the temperature of a body
changes at a rate proportional to the difference in temperature
between its own temperature and the temperature of its
surroundings.
We can therefore write
dT
= −k (T − Ts )
dt
Dr. Marco A Roque Sol
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where,
T = Temperature of the body at any time t
Ts = Temperature of the surroundings (also called ambient
temperature)
T0 = Initial temperature of the body
k = constant of proportionality
Integrating this separable differential equation
dT
= −k (T − Ts )
dt
Dr. Marco A Roque Sol
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1
dT = −kdt
T − Ts
Z
Z
1
dT = − kdt
T − Ts
ln|T − Ts | = −kt + C
|T − Ts | = e −kt+C
Dr. Marco A Roque Sol
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T = Ts + ce −kt ;
T > Ts
(why ?)
Applying initial conditions T (0) = T0
c = T0 − Ts
Thus, the solution is
T = Ts + (T0 − Ts )e −kt
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
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Linear and Nonlinear Equations
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Theorem 1 (Existence and Uniqueness).
Suppose p(t) and g (t) are continuous real-valued functions on an
interval (α, β) that contains the point t0 . Then, for any choice of
(initial value) y0 , there exists a unique solution y (t) on the whole
interval (α, β) to the linear differential equation
dy (t)
+ p(t)y (t) = g (t)
dt
for all t ∈ (α, β) and y (t0 ) = yo.
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
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Linear and Nonlinear Equations
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Theorem 2 (Existence and Uniqueness).
Suppose that both f (t, y ) and
∂f
(t, y ) are continuous functions
∂y
defined on a region R as
R = {(t, y ) : t0 − δ < t < t0 + δ; y0 − < y < y0 + }
containing the point (t0 , y0 ). Then there exists a number δ1
(possibly smaller than δ) so that a solution y = f (t) to
y 0 = f (t, y )
y (t0 ) = y0
is the unique solution for t0 − δ1 < t < t0 + δ1
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Modeling
Linear and Nonlinear Equations
Autonomous Equations
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Linear and Nonlinear Equations
Example 21
Consider the ODE
y 0 = t − y + 1;
y (1) = 2
In this case, both the function f (t, y ) = ty + 1 and its partial
∂f
derivative
(t, y ) = −1 are defined and continuous at all points
∂y
(x, y ). The theorem guarantees that a solution to the ODE exists
in some open interval centered at 1, and that this solution is
unique in some (possibly smaller) interval centered at 1
Dr. Marco A Roque Sol
Ordinary Differential Equations
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In fact, an explicit solution to this equation is
y (t) = t + e 1t
This solution exists (and is the unique solution to the equation) for
all real numbers t. In other words, in this example we may choose
the number δ and δ1 as large as we please.
Example 22
Consider the ODE
y 0 = 1 + y 2;
Dr. Marco A Roque Sol
y (0) = 0
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In this case, both the function f (t, y ) = 1 + y 2 and its partial
∂f
derivative
(t, y ) = 2y are defined and continuous at all points
∂y
(t, y ).
The theorem guarantees that a solution to the ODE exists in some
open interval centered at 0, and that this solution is unique in
some (possibly smaller) interval centered at 1
By separating variables and integrating, we derive a solution to this
equation of the form
y (t) = tan(t)
Dr. Marco A Roque Sol
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As an abstract function of t, this is defined for all
t 6= ..., −3π/2, π/2, π/2, 3π/2, .... However, in order for this
function to be considered as a solution to this ODE, we must
restrict the domain. Specifically, the function
y (t) = tan(t);
−π/2 < x < π/2
is a solution to the above ODE. In this example we must choose
δ1 = π/2, although the initial value δ, may be chosen as large as
we please.
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
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Linear and Nonlinear Equations
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One particular case of a separable equation is the so called
Autonomous Equation, which is an equation of the form
dy
= f (y )
dt
That is, the right hand side only depends on y and therefore the
directional field can be found only analyzing horizontal lines.
We will use some concrete examples to see how we can use a
qualitative analysis to determine the behavior of the function.
Dr. Marco A Roque Sol
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Population
If P(t) represents a population in a given region at any time t the
basic equation that we will use is identical to the one that we used
for mixing. Namely,
Rate of change of P(t) = Rate at which P(t) enters the region
Rate at which P(t) exits the region
Here the rate of change of P(t) is still the derivative. For
population problems all the ways for a population to enter the
region are included in the entering rate. Birth rate and migration
into the region are examples of terms that would go into the rate
at which the population enters the region.
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
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Likewise, all the ways for a population to leave an area will be
included in the exiting rate. Therefore things like death rate,
migration out and predation are examples of terms that would go
into the rate at which the population exits the area.
Example 23
A population of insects in a region will grow at a rate that is
proportional to their current population. In the absence of any
outside factors the population will triple in two weeks time. On
any given day there is a net migration into the area of 15 insects
and 16 are eaten by the local bird population and 7 die of natural
causes. If there are initially 100 insects in the area will the
population survive? If not, when do they die out?
Dr. Marco A Roque Sol
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Solution
Let’s start out by looking at the birth rate. We are told that the
insects will be born at a rate that is proportional to the current
population. This means that the birth rate can be written as
rP(t)
where r is a positive constant that will need to be determined.
Now, let’s take everything into account and get the IVP for this
problem.
P 0 (t) = (rP + 15) − (16 + 7)
P(0) = 100
P 0 (t) = rP − 8 P(0) = 100
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
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Note that we don’t make use of the fact that the population will
triple in two weeks time in the absence of outside factors here. In
the absence of outside factors means that the only thing that we
can consider is birth rate. Nothing else can enter into the picture
and clearly we have other influences in the differential equation.
So, just how does this tripling come into play? We will use the fact
that the population triples in two week time to help us find r . In
the absence of outside factors the differential equation would
become.
P 0 (t) = rP
P(0) = 100
P(14) = 300
Dr. Marco A Roque Sol
Ordinary Differential Equations
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This differential equation is separable and linear and is a simple
differential equation to solve. The general solution is :
P(t) = ce rt
Applying the initial condition gives c = 100. Now apply the second
condition.
300 = P(14) = 100e 14r
And solving for r , we have
r=
ln3
14
Dr. Marco A Roque Sol
Ordinary Differential Equations
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Now, that we have r we can go back and solve the original
differential equation. We will rewrite it a little for the solution
process.
ln3
0
P = −8 P(0) = 1000
P (t) −
14
This is a simple linear differential equation, with integrating fractor
given by:
µ(t) = e −
R
ln3
dt
14
ln3
= e − 14 t
And the solution is
Z Z
0
ln3
− ln3
t
14
Pe
dt = −8 e − 14 t dt
Dr. Marco A Roque Sol
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P(t) =
ln3
112
+ ce 14 t
ln3
And applying the initial condition, we get
ln3
112
112 ln3 t
112
P(t) =
+ 100 −
e 14
=
− (1.947) e 14 t
ln3
3
ln3
Now, the exponential has a positive exponent and so will go to
plus infinity as t increases. Its coefficient, however, is negative and
so the whole population will go negative eventually. Since the
population cannot be negative, there is a time such tha,t all of
them die out
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
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ln3
112
− (1.9468) e 14 t = 0
ln3
t = 50.4 days
Thus, the insects will survive for around 7 weeks.
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Modeling
Linear and Nonlinear Equations
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Example 24
Logistic Growth
Previously we have modeled a population based on the assumption
that the growth rate would be a constant
A much more realistic model of a population growth is given by
the logistic growth equation. Here is the logistic growth equation.
P
0
P =r 1−
P
K
In the logistic growth equation r is the intrinsic growth rate. It is
the growth rate that will occur in the absence of any limiting
factors. K is called either the saturation level or the carrying
capacity.
Dr. Marco A Roque Sol
Ordinary Differential Equations
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Now, we claimed that this was a more realistic model for a
population. Lets see if that in fact is correct. To allow us to sketch
a direction field let’s pick a couple of numbers for r and K . We will
use r = 1/2 and K = 10. For these values the logistics equation is.
P
1
0
1−
P
P =
2
10
First notice that the derivative will be zero at P = 0 and P = 10.
Also notice that these are in fact solutions to the differential
equation. These two values are called equilibrium solutions since
they are constant solutions to the differential equation. Here is the
direction field as well as some solutions sketched in as well.
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Modeling
Linear and Nonlinear Equations
Autonomous Equations
Exact Equations
Autonomous Equations
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
Modeling
Linear and Nonlinear Equations
Autonomous Equations
Exact Equations
Autonomous Equations
Now, lets move on to the point of this section. The logistics
equation is an example of an autonomous differential equation.
dy
= f (y )
dt
Notice that if f (y0 ) = 0 for some value y = y0 ) = 0 then this will
also be a solution to the differential equation. These values are
called equilibrium solutions or equilibrium points. What we
would like to do is classify these solutions.
So, let’sgo backto our logistic equation
1
P
P0 =
1−
P
2
10
Dr. Marco A Roque Sol
Ordinary Differential Equations
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As we pointed out there are two equilibrium solutions to this
equation P = 0 and P = 10. If we the fact that we’re dealing with
population these points break up the P number line into two
distinct regions.
0 < P < 10;
10 < P < ∞
We will say that a solution starts near an equilibrium solution if it
starts in a region that is on either side of that equilibrium solution.
Equilibrium solutions, in which solutions that start near them move
away from the equilibrium solution, are called unstable
equilibrium points or unstable equilibrium solutions. So, for
our logistics equation, P = 0 is an unstable equilibrium solution.
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
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Linear and Nonlinear Equations
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Next, solutions that start near P = 10 all move in toward P = 10
as t increases. Equilibrium solutions in which solutions that start
near them move toward the equilibrium solution are called
asymptotically stable equilibrium points or asymptotically
stable equilibrium solutions. So, P = 10 is an asymptotically
stable equilibrium solution
Example 25
Find and classify all the equilibrium solutions to the following
differential equation.
y0 = y2 − y − 6
Dr. Marco A Roque Sol
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Solution
First, find the equilibrium solutions. This is generally easy enough
to do.
y 2 − y − 6 = (y − 3)(y + 2) = 0
So, it looks like we’ve got two equilibrium solutions. Both y = −2
and y = 3 are equilibrium solutions. Thus, those points divide the
pane in three sections, namely
−∞ < y < −2;
−2 < y < 3;
3<y <∞
In those regiones we have the following behavior for y 0 :
Dr. Marco A Roque Sol
Ordinary Differential Equations
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In the region −∞ < y < −2 the derivative y 0 is positive.
In the region −2 < y < 3 the derivative y 0 is negative.
In the region 3 < y < ∞ the derivative y 0 is positive.
Thus, y = −2 is an asymptotically stable equilibrium solution and
y = 3 is an unstable equilibrium solution.
Example 26
Find and classify all the equilibrium solutions to the following
differential equation.
y 0 = (y 2 − 4)(y + 1)2
Dr. Marco A Roque Sol
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Solution
First, find the equilibrium solutions. This is generally easy enough
to do.
(y 2 − 4)(y + 1)2 = (y − 2)(y + 2)(y + 1)2 = 0
So, it looks like we’ve got three equilibrium solutions. Both
y = −2, y = −1 and y = 2 are equilibrium solutions. Thus, those
points divide the pane in four sections, namely
−∞ < y < −2;
−2 < y < −1;
−1 < y < 2;
2<y <∞
In those regiones we have the following behavior for y 0 :
Dr. Marco A Roque Sol
Ordinary Differential Equations
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In the region −∞ < y < −2 the derivative y 0 is positive.
In the region −2 < y < −1 the derivative y 0 is negative.
In the region −1 < y < 2 the derivative y 0 is negative again.
In the region 2 < y < ∞ the derivative y 0 is positive.
Thus, y = −2 is an asymptotically stable equilibrium solution,
y = −1 is a semi-stable equilibrium solution, and y = 2 is an
unstable equilibrium solution.
Dr. Marco A Roque Sol
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The next type of first order differential equations that we will be
looking at is exact differential equations. Suppose that we have
the following differential equation.
M(x, y ) + N(x, y )
dy
=0
dx
or equivalently
M(x, y )dx + N(x, y )dy = 0
Now, if there is a function somewhere out there in the world,
ψ(x, y ), so that
Dr. Marco A Roque Sol
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∂ψ(x, y )
=M
∂x
and
∂ψ(x, y )
=N
∂y
then we call the differential equation exact. In these cases we can
write the differential equation as
∂ψ(x, y ) ∂ψ(x, y ) dy
+
=0
∂x
∂y
dx
Dr. Marco A Roque Sol
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Then using the chain rule from Calculus III we can further reduce
the differential equation to the following derivative:
d
(ψ (x, y (x))) = 0
dx
this implies that
ψ (x, y (x)) = c = constant
This then is an implicit solution for our differential equation!
Dr. Marco A Roque Sol
Ordinary Differential Equations
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Finding the function ψ(x, y ) is clearly the central task in
determining if a differential equation is exact and in finding its
solution. Therefore, it would be nice if there was some simple test
that we could use before even starting to see if a differential
equation is exact or not.
Since it’s exact we know that somewhere out there is a function
ψ(x, y ) that satisfies
∂ψ(x, y )
=M
∂x
Dr. Marco A Roque Sol
and
∂ψ(x, y )
=N
∂y
Ordinary Differential Equations
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Now, provided ψ(x, y ) is continuous and its first order derivatives
are also continuous we know that
ψ(x, y )xy = ψ(x, y )yx
However, we also have the following.
ψ(x, y )xy = (ψ(x, y )x )y = (M)y = My
ψ(x, y )yx = ψ(x, y )y = (N)x = Nx
x
Therefore, if a differential equation is exact and ψ(x, y ) meets all
of its continuity conditions we must have.
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
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M y = Nx
Therefore, we will use the above equation as a test for exact
differential equations. If so, we will assume that the differential
equation is exact and that ψ(x, y ) meets all of its continuity
conditions and proceed with finding it
Example 27
Solve the following IVP .
2xy − 9x 2 + (2y + x 2 + 1)
Dr. Marco A Roque Sol
dy
= 0;
dx
y (0) = −3
Ordinary Differential Equations
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Solution
First identify M and N and check that the differential equation is
exact.
M = 2xy − 9x 2
N = 2y + x 2 + 1
My = 2x = Nx
So, the differential equation is exact according to the test. Now,
how do we actually find ψ(x, y ) ? Well recall that
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
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Linear and Nonlinear Equations
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ψ(x, y )x = M
ψ(x, y )y = N
We can use either of these to get a start on finding ψ(x, y ) by
integrating as follows.
Z
Z
ψ(x, y ) = Mdx + h(y ) or ψ(x, y ) = Ndy + g (x)
Often it doesn’t matter which one you choose to work with while
in other problems one will be significantly easier than the other.
Dr. Marco A Roque Sol
Ordinary Differential Equations
First Order Differential Equations
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So, I will use the first one.
Z
ψ(x, y ) = (2xy − 9x 2 )dx = x 2 y − 3x 3 + h(y )
Okay, we have got most of the function ψ(x, y ), we just need to
determine h(y ) and we will be done. Differentiate our ψ(x, y ) with
respect to y and set this equal to N :
ψ(x, y )y = x 2 + h0 (y ) = N = 2y + x 2 + 1
From this we can see that
h0 (y ) = 2y + 1
Dr. Marco A Roque Sol
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Note that at this stage h(y ) must be only a function of y and so if
there are any x 0 s in the equation at this stage something is wrong
and it’s time to go look for it.
We can now find h(y ) by integrating.
h(y ) = y 2 + y + k
So, we can now write down ψ(x, y )
ψ(x, y ) = x 2 y − 3x 3 + y 2 + y + k
implicit solution is then
ψ(x, y ) = C =⇒ y 2 + (x 2 + 1)y − 3x 3 = C − k = c
Dr. Marco A Roque Sol
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Applying the initial conditions
(−3)2 + ((0)2 + 1)(−3) − 3(0)3 = c
=⇒ c = 6
The complete (including the constant)implicit solution is then.
y 2 + (x 2 + 1)y − 3x 3 − 6 = 0
Now, this is quadratic in y and so we can solve for y (x) by using
the quadratic formula.
p
−(x 2 + 1) ± (x 2 + 1)2 − 4(1)(−3x 3 − 6)
y=
2(1)
Dr. Marco A Roque Sol
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p
(x 4 + 12x 3 + 2x 2 + 25
y=
2
Now, reapply the initial condition to figure out which of the two
signs in the that we need.
√
−1 ± 25
−1 ± 5
−3 = y (0) =
=
= −3, 2
2
2
So, it looks like the minus sign is the one that we need. The
explicit solution is then.
p
−(x 2 + 1) − (x 4 + 12x 3 + 2x 2 + 25
y=
2
−(x 2 + 1) ±
Dr. Marco A Roque Sol
Ordinary Differential Equations