Introduction
First Order Differential Equations
Ordinary Differential Equations
Dr. Marco A Roque Sol
12/01/2015
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Direction Fields
Introduction
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Direction Fields
Direction Fields
dv
γv
=g−
dt
m
In order to look at direction fields, it would be helpful to have some
numbers for the various quantities in the differential equation.
So, let’s assume that we have a mass of 2 kg and that γ = 0.392.
Plugging this into above gives the following differential equation.
dv
= 9.8 − 0.196v
dt
whose direction field is given by
m
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Direction Fields
Direction Fields
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Direction Fields
Direction Fields
So, just why do we care about direction fields? There are two nice
pieces of information that can be readily found from the direction
field for a differential equation.
Sketch of solutions. Since the arrows in the direction fields are in
fact tangents to the actual solutions to the differential equations
we can use these as guides to sketch the graphs of solutions to the
differential equation.
Long Term Behavior. In many cases we are interested in how the
solutions behave as t increases. Direction fields, if we can get our
hands on them, can be used to find information about this long
term behavior of the solution.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Direction Fields
Direction Fields
Thus, in the case of a falling body we have that the long-term
behavior can be seen from
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Direction Fields
Direction Fields
dy
Another example is
= (y 2 − y − 2)(1 − y )2 . Whose direction
dt
field is
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Direction Fields
Direction Fields
An solutions are given by
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Direction Fields
Direction Fields
dy
Finally, we have the next example
= e t − 2y . Whose direction
dt
field and some solutions are
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
First Order Differential Equations
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
The first special case of first order differential equations that we
will look is the linear first order differential equation.
a(t)y 0 + b(t)y = c(t)
with a(t), b(t), and c(t) continuous function on some interval.
Example 1
(4 + t 2 )
dy
+ 2ty = 4t
dt
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
Solution
d
(4 + t 2 )y = 4t
dt
Z
2
(4 + t )y = 4tdt + C ; C = constant
(4 + t 2 )y = 2t 2 + C
y=
2t 2
C
+
2
(4 + t ) (4 + t 2 )
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
In order to solve a linear first order differential equation we MUST
start with the differential equation in the form shown below:
dy
+ p(t)y = g (t)
dt
Where both p(t) and g (t) are continuous functions.
If the differential equation is not in this form then the process we
are going to use will not work.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
Now, we are going to assume that there is some function
somewhere out there, µ(t), called an integrating factor, such
that, when we multiply the equation by it :
dy
+ µ(t)p(t)y = µ(t)g (t)
dt
that function has the property that
µ(t)
µ(t)p(t) = µ0 (t)
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
And in this way the ODE becomes
dy
+ µ0 (t)y = µ(t)g (t)
dt
and using the product rule we obtain
µ(t)
(µ(t)y (t))0 = µ(t)g (t)
Integrating and solving for y , we have
Z
1
y (t) =
µ(t)g (t)dt + C ;
µ
Dr. Marco A Roque Sol
C = constant
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
So, now that weve got a general solution to a linear first order we
need to go back and determine just what this function µ(t) is.
This is actually an easier process. We will start with
µ0 (t) = µ(t)p(t)
µ0 (t)
= p(t)
µ(t)
ln (µ(t))0 = p(t)
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
Integrating, we obtain
Z
lnµ(t) = p(t)dt + K
Exponentiate both sides to get µ(t) out of the natural logarithm.
µ(t) = e
R
p(t)dt+K
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
Now, let’s make use of the fact that K is an unknown constant. If
K is an unknown constant then so is e K , so we might as well just
rename it k and make our life easier. This will give us the
following.
µ(t) = ke
R
p(t)dt
So, we now have a formula for the general solution of the linear
first order differential equation, and a formula for the integrating
factor.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
We do have a problem however. We’ve got two unknown constants
and the more unknown constants we have the more trouble we will
have later on. Therefore, it would be nice if we could find a way to
eliminate one of them (we will not be able to eliminate both ???).
This is actually quite easy to do:
Z
R
1
y (t) = R p(t)dt
ke p(t)dt g (t)dt + C ; C = constant
ke
Z R
1
y (t) = R p(t)dt k e p(t)dt g (t)dt + C ; C = constant
ke
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
y (t) =
y (t) =
1
e
R
e
R
p(t)dt
1
p(t)dt
C
g (t)dt +
; C = constant
e
k
Z R
p(t)dt
g (t)dt + c ; c = constant
e
Z
R
p(t)dt
The solution to a linear first order differential equation is then
Z
1
y (t) =
µ(t)g (t)dt + c ; c = constant
µ
where
µ(t) = e
R
p(t)dt
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
Now, we will need to use the integratin factor µ(t) regularly, as
that formula is easier to use than the process to derive it.
Solution Process
The solution process for a first order linear differential equation is
as follows.
1.- Put the linear differential equation in the correct initial form.
dy
+ p(t)y (t) = g (t)
dt
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
2.-Find the integrating factor µ(t).
µ(t) = e
R
p(t)dt
3.- Multiply everything in the differential equation by µ(t) and
verify that the left side becomes the product rule (µ(t)y (t))0 and
write it as such.
4.- Integrate both sides, make sure you properly deal with the
constant of integration.
5.- Solve for the solution y (t).
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
Example 2
Find the solution to the following differential equation:
dv
= 9.8 − 0.196v
dt
First we need to get the differential equation in the correct form:
dv
+ 0.196v (t) = 9.8
dt
From this we can see that p(t) = 0.196 and so µ(t) is then.
µ(t) = e
R
0.196dt
Dr. Marco A Roque Sol
= e 0.196t
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
Multiply everything in the differential equation by µ(t) :
e 0.196t
dv
+ e 0.196t 0.196v (t) = 9.8e 0.196t
dt
0
e 0.196t v (t) = 9.8e 0.196t
Integrate both sides, make sure you properly deal with the
constant of integration.
Z
Z
0
0.196t
e
v (t) dt = 9.8e 0.196t dt
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
e
0.196t
Z
v (t) =
9.8e 0.196t dt
and integrating
e 0.196t v (t) =
9.8 0.196t
e
+c
0.196
Solve for the solution y (t).
v (t) = 50 + ce −0.196t
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
From the solution to this example we can now see why the
constant of integration is so important in this process.
Without it, in this case, we would get a single, constant solution,
v (t) = 50.
With the constant of integration we get infinitely many solutions,
one for each value of c.
To sketch some solutions, all we need to do is to pick different
values of c . Several of these are shown in the graph below:
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
Now, recall from the Definitions section that the Initial
Condition(s) will allow us to pick particular solution.
Solutions to first order differential equations (not just linear as we
will see) will have a single unknown constant in them and so we
will need exactly one initial condition to find the value of that
constant and hence find the solution that we were after.
The initial condition for first order differential equations will be of
the form
y (t0 ) = y0
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
Recall as well, that a differential equation along with a sufficient
number of initial conditions is called an Initial Value Problem
(IVP) .
Example 3
Solve the following IVP
dv
= 9.8 − 0.196v , v (0) = 48
dt
To find the solution to an IVP, we must first find the general
solution to the differential equation and then use the initial
condition to identify the exact solution that we are after.
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
So, since this is the same differential equation as we looked at in
Example 2, we already have its general solution.
v (t) = 50 + ce −0.196t
And applying the initial conditions we have
48 = 50 + ce (−0.196)(0)
48 = 50 + c
c = −2
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
Dr. Marco A Roque Sol
Ordinary Differential Equations
Introduction
First Order Differential Equations
Linear Equations
Linear Equations
So, the actual solution to the IVP is:
v (t) = 50 − 2e (−0.196)(t)
Dr. Marco A Roque Sol
Ordinary Differential Equations