This chapter corresponds roughly to the combined Chapters 1 and 2 in the textbook, although there
are some differences in the order of presentation and some content differences as well.
Table of Contents:
1.0 Introduction to Differential Equations – Introduction and Generalities
1.1 Introduction to Differential Equations – Terminology and classification
1.2 Direct integration for equations of the form y ( n ) ( x)  f ( x)
1.3 Slope Fields and Solution curves of y '  f ( x, y )
1.4 Linear first order equations y ' p( x) y( x)  q( x) and applications
1.5 Separable equations 𝑦 ′ = 𝑃(𝑥) ∗ 𝑄(𝑦)
1.6 Exact equations
1.7 Substitution methods
1.8 Existence and uniqueness for first order odes y '  f ( x, y )
1.9 Mathematical models: population models and velocity acceleration models
1
1.0 Introduction to Differential Equations – Introduction and Generalities
In this Introduction to Differential Equations class we’ll learn:
 to solve fundamental types of differential equations which lays a solid basis for solving differential equations
in general;
 to interpret the solutions found and to use other graphs and methods (such as direction fields) to obtain
(properties of a ) solution;
 to see the relation between these differential equations (de’s) and practical problems .
Many of the physical principles of nature involve rates at which things happen and quantities change (e.g. the motion
of an object or of a body, the rate of change of a process such as the increase of a volume of liquid in a tank, the
variation of a certain population, the motion of fluids, the flow of current in an electrical circuit, etc). When expressed
in mathematical terms, these processes produce equations that contain derivatives, which we call differential
equations. For this reason, de’s are often present in mathematical models of physical processes.
The usual steps of analyzing a mathematical model are:
 formulate the model, i.e. identify the quantities of interest, assign letters to them, and express the physical
principles in terms of rates of change of these variables, that is find the fundamental differential equation(s) of
the model. We may also need to think of initial conditions and/or other statical conditions of the system. In
many of these models, time is often the independent variable;
 Solve the differential equation/ initial (boundary) value problem (ivp/bvp);
 Interpret the solution to generate the solution and in consequence the property of interest of your solution (the
solution at a certain later time or in the long run, etc).
We will cover the following topics according to our syllabus:
1. First order differential equations 𝑦 ′ = 𝑓(𝑥, 𝑦)
2. Second order linear differential equations: 𝐴(𝑥)𝑦 ′′ + 𝐵(𝑥)𝑦 ′ + 𝐶(𝑥)𝑦 = 𝑓(𝑥) .
3. Higher order linear differential equations: 𝐴𝑛 (𝑥)𝑦 (𝑛) + 𝐴𝑛−1 (𝑥)𝑦 (𝑛−1) +  + 𝐴1 (𝑥)𝑦 ′ + 𝐴0 (𝑥) = 𝑓(𝑥)
4. Power series methods for linear differential equations with variable coefficients
5. Linear systems of differential equations: 𝑋 ′ = 𝐴𝑋
6. Nonlinear differential equations and stability (if time permits).
2
Examples of mathematical models that produce differential equations (see also textbook):
a) Newton’s law of cooling which models the temperature T of an object submerged into an environment of a
different constant temperature A: 𝑑𝑇
=−𝑘(𝑇−𝐴), where k is a positive constant.
𝑑𝑡
b) Population models:
i) The simplest model, assuming constant no external factors:
𝑑𝑃
𝑑𝑡
= 𝑘𝑃, i.e the rate of change of the
population is proportional with the current population. This assumes that we have no external effects and
infinite resources and constant birth and death rates.
ii) Considering limited resources and presence of predators and possible linear birth and/or death rates, there
is a limit L in the population size. Therefore, a better model would be:
𝑑𝑃
𝑑𝑡
= 𝑘(𝐿 − 𝑃)𝑃 – logistic model.
c) Motion of an object due to gravitational acceleration, such as free fall or other vertical motion:
𝑚𝑥 ′′ = −𝑚𝑔
(here we ignore any other effects than gravitation and consider that x(t) increases upwards). Better models
would include other effects, such as air friction.
 mx ''  0
Similarly, a projectile motion: 
with typical initial conditions:
my ''  mg
v(0)   v0 cos  , v0 sin  

x(0)   x0 , y0 

where v0 is the magnitude of the initial velocity (a vector that makes an angle  with the x axis).
3
1.1 Introduction to Differential Equations – Terminology and classification
In this section we’ll learn:




What is a differential equation;
Get some idea about the methods to solve differential equations;
Classification of differential equations;
Examples of typical differential equations.
A. What is a differential equation (de):
Definition 1.1.1: A differential equation is an equation which contains derivatives (of an unknown function which
needs to be found).
Examples:
1) 𝑦 ′ = 𝑘𝑦 (as in population models)
2) 𝑦 ′ = 2𝑥𝑦
3) 𝑦 ′ = tan⁡(𝑥) (this is an example of de simple to solve by a method called direct integration)
4) 𝑦 ′ = 𝑦 2 (particular logistic model)
5) 𝑦 ′′ + 9𝑦 = 0 (mechanical mass-spring vibrations)
6) 𝑦 ′′ − 𝑥𝑦 = 0 (Airy’s function, solution of Schrodinger equation)
7) 𝑦 ′′ + ln⁡(𝑥)𝑦 = 0
8) 𝑦 ′′ + 𝑒 𝑦 = 0
𝑑𝑦
(in all these de’s 𝑦′ ≡ 𝑑𝑥 , assuming that x is the independent variable).
B. Classification of differential equations:
I) By order:
Definition 1.1.2: The order of a differential equation is the order of the highest derivative present in the equation
(note that some de’s may not be organized/solved by the highest derivative and you have to determine the order
carefully).
In our examples, 1)4) are first order de’s and 5) 8) are second order de’s .
4
A general n’th order ordinary differential equation has the form:
(1) 𝐹(𝑥, 𝑦(𝑥), 𝑦 ′ (𝑥), 𝑦 ′′ (𝑥), ⁡, 𝑦 (𝑛) (𝑥)) = 0
Examples:
1) (𝑦′)2 + 𝑦 2 = 0
2) 𝑦 (4) + 𝑥 2 𝑦 (3) + 𝑥 5 𝑦 = sin⁡(𝑥) (fourth order ordinary de).
However, de’s are usually given explicitely (not implicitely) in terms of 𝑦 (𝑛) (𝑥) (this is an algebraic step which is
usually performed before solving the equation).
A general explicit n’th order de has the form:
(2) 𝑦 (𝑛) (𝑥) = 𝑓(𝑥, 𝑦(𝑥), 𝑦 ′ (𝑥), 𝑦 ′′ (𝑥), ⁡, 𝑦 (𝑛−1) (𝑥))
In particular, of fundamental significance is the explicit first order ode of the form:
(3) 𝑦 ′ = 𝑓(𝑥, 𝑦)
Equation (3) will be our subject of study for the first Chapter.
II) By the number of independent variables:
 For most of this class we’ll be solving differential equations for which the unknown function depends on only
ONE independent variable. These differential equation (with only one independent variable) are called
ordinary differential equations (ode’s).
 Ocassionally, we may also look at differential equations for which the unknown function depends on two or
more independent variables, and then we will have partial derivatives, which denote the rate of change of the
unknown function with respect to one variable while all the others are assumed fixed. These are called partial
differential equations (pdes).
For example:
1)
𝑢
𝑡
2 𝑢
= 𝑘 𝑡 2 where 𝑢 = 𝑢(𝑡, 𝑥) -- heat equation (second order partial differential equation).
III) Based on the form of the equation, we distinguish between linear and non-linear de’s:
Definition 1.1.3: A general n’th order ode 𝐹(𝑥, 𝑦(𝑥), 𝑦 ′ (𝑥), 𝑦 ′′ (𝑥), ⁡, 𝑦 (𝑛) (𝑥)) = 0 is a linear ode if the function F
is linear with respect to y(x) and its derivatives, but not necessarily with respect to the independent variable.
Therefore, a n’th order linear ode has the form:
(4) 𝑎0 (𝑥)𝑦(𝑥) + 𝑎1 (𝑥)𝑦 ′ (𝑥) + 𝑎2 (𝑥)𝑦 ′′ (𝑥) +  + 𝑎𝑛 (𝑥)𝑦 (𝑛) (𝑥) = 𝑔(𝑥)
5
Note: In (4), 𝑎0 (𝑥), , 𝑎𝑛 (𝑥), 𝑔(𝑥) can very well be non-linear functions, but y(x) and its derivatives must appear in
a linear combination. Another way to decide if an ode is linear is (other than comparing with (3)) by verifying that the
lhs of (4) has the property 𝑙ℎ𝑠(𝑦1 (𝑥) + 𝑦2 (𝑥)) = 𝑙ℎ𝑠(𝑦1 (𝑥)) + 𝑙ℎ𝑠(𝑦2 (𝑥)).
Otherwise, if the de is not of the form (4), that is is F is nonlinear in y(x) and/or its derivatives, the ode is called nonlinear.
Examples:
1
1) 𝑦 ′′ + 𝑥 𝑦 ′ + 𝑦 = 0, 𝑥 > 0 -- linear second order ordinary differential equation
2) 𝑦 ′′ + sin⁡(𝑥)𝑦 ′ = 0 -- linear second order ordinary differential equation
3) 𝑦′′′ + y𝑦 ′ = 0 -- non-linear third order ode
4) 𝑦 ′′ + sin⁡(𝑦) = 0 -- nonlinear second order ode
5) 𝑦′ = 𝑦 2 -- nonlinear first order ode
Therefore, in a linear ode, the function y(x) and its derivatives appear alone (eventually multiplied by some function
a(x)), and they are not multiplied one with another. A non-linear ode permits non-linearities in y(x) and its
derivatives. This distinction linear-nonlinear is very important and we should make sure that we understand it well.
IV) By the number of de’s , we can have :
 Scalar des (only one differential equation to solve with one unknown function): (e.g. 𝑦 ′ + 𝑦 = 0 ).
 System of differential equations, where we have 2 (or more) differential equations with 2 (or more) unknowns:
𝑑𝑥
𝑑𝑡
Example: 1) {𝑑𝑦
𝑑𝑡
= 3𝑥 + 𝑦
= −𝑥 + 𝑦
For most of our course, we’ll be concerned with scalar ordinary differential equations or first, second, and in topic 4
higher order.
AS mentioned, Topic 2 (Chapter 1) will be concerned with scalar first order (linear or nonlinear) ordinary differential
equations of the general form (3). This is a fundamental differential equation which needs to be known well, the next
types of ode’s often relate to this. Therefore, our goal for the beginning is studying equations of form (2) or (3),
usually of first and second order.
6
Consider now the typical general equation (2).
Definition 1.1.4: A function 𝜑 = 𝜑(𝑥) is a solution of (2) 𝑦 (𝑛) (𝑥) = 𝑓(𝑥, 𝑦(𝑥), 𝑦 ′ (𝑥), 𝑦 ′′ (𝑥), ⁡, 𝑦 (𝑛−1) (𝑥)) for
x  I if  ( x) is n times differentiable on I (that is  ( x) is continuous on I and all derivatives  '( x),  ''( x),
,  ( n ) ( x)
exist on I) and if (𝑛) (𝑥) = 𝑓(𝑥, (𝑥), ′ (𝑥), ′′ (𝑥), ⁡, (𝑛−1) (𝑥)) .
Examples:
1) Verify that y ' y  0 has a (general) solution of the form y ( x)  Ce x for   x  
2) Verify that y( x)  2 x1/2  x1/2 ln( x) satisfies the differential equation 4 x 2 y '' y  0 for x  0 . (Example 8/
1.1)
3) Show that the second order ode y '' y  0 has a (particular) solution of the form
y( x)  cos  x  . What other solutions of this differential equation can we find?
In principle, a general solution of a differential equation is a solution (formula) describing all possible solutions of
that equation (often containing one or more arbitrary constants of integration – see example 1) above). This is typical
for differential equations if no extra (initial or boundary conditions) are posed – after all, the de describes just a rate of
change. A solution free from arbitrary constants (perhaps chosen to satisfy certain initial / boundary conditions) is
called a particular solution of a differential equation.
Example:
1) The differential equation y ' y  1 has y  1 as a particular solution, and yG ( x)  Ce  x  1 as a general
solution. Find the arbitrary constant C such that yG ( x) also satisfies the initial condition y (0)  2 .
Important questions:
Although for the de’s above we were able to verify that a certain function is a solution of that de, in general we do not
have such solutions readily available. Therefore, a fundamental question is: Does an equation of the form (2) always
have a solution? The answer is “No”. Just by writing the equation it does not mean that there is a function y   ( x )
that satisfies it. This is the question of existence of a solution, and it is answered by a theorem (we’ll cover it a bit
later) stating that under certain restrictions on the function f , the equation (2) is guaranteed to have solutions.
The next question is that, if we determine that (2) has solutions, how many solutions does it have? This is the question
of uniqueness. In general, solutions of differential equations contain one or more constant of integration, as above.
However, if one (or more, depending on the order of the equation) extra initial/boundary conditions are imposed,
these constant will be determined/eliminated. Therefore, an uniqueness theorem (also covered later) will guarantee
that under certain conditions on f , then there is a unique solution of a given initial value problem.
A third question is, given a differential equation of the form (2), can we always determine a solution, and if so, how?
For simple differential equations (Topics 2,3, and 5) we’ll be able to determine a solution in terms of elementary
functions (polynomial, exponential, trigonometric, logarithmic or hyperbolic functions). However, for more general
(most) differential equations, this is not possible, and methods of a more general nature that seek approximations of
solutions are introduced (Topic 4, 6).
Homework: Problems 1,6,9,11,14,15,19,21,24,27,28,31,32,40 and 41 from Section 1.1 in the Textbook.
7
1.2 Direct integration for equations of the form y ( n ) ( x)  f ( x)
For the next few sections, we’ll learn specific methods for solving particular types of first order ode’s of the
general form (1) y '  f  x, y  .
In this section, we’ll be looking at methods to solve the simpler particular form of (1): (2) y '  f ( x) , as well
as applications of this equation. We call (2) (and its higher order general form to be presented soon) the direct
integration equation. This is because (2) can be solved easily by integrating both sides, which produces:
(3) y   f ( x)dx  C  F ( x)  C , where F ( x) is an anti-derivative of f ( x) , as long as we know how to calculate this
anti-derivative. (remember that
 y ' dx  y  C ).
If we also have an initial condition y  x0   y0 , we use it in (3) to find C  y0  F ( x0 ) .
(3) represents the general solution of (2).
Example: Solve the initial value problem (ivp): (4) y ' 
x
x
2
 9
1/2
, y (4)  2 .
Since this is a direct integration ode (compare with (2)), integrate both sides to find
(5) y  
x
x
2
 9
1/2
dx 
  x2  9
1/2
 C  x2  9  C .
All of these are solutions of (4). When plotting these (Mathematica) we have the solutions of (4) which are here
parallel (they all verify (4) in the sense that they have the same rate of change (when compared with one another) for
all x).
12
10
8
6
4
2
10
5
5
10
Figure 1.2.1: Mathematica generated graph of (5) for C=-3,-2,-1,0,1,2
8
To find the particular solution which verifies the ic in (4) , impose this in (5) and find C=-3.
Similarly we can solve a second order direct integration equation of the form:
(6) y ''  f ( x) or an n-th order direct integration equation of the form (7) y ( n )  f ( x) .
For example, integrating (6) once we find: y '   f ( x)dx  C1  F ( x )  C1 and integrating once again we find:
y   F ( x)dx  C1 x  C2 , where C1 and C2 are constant of integration (note that a second order ode will have in
general two constants of integration and an n-th order ode has in general n constants of integration).
Application: Velocity and Acceleration:
Problems from mechanics (Newton’s second law) are typical direct integration problems, since we know the force
acting on an object and we want to find the object’s position.
In this context, a particle is described by its position (on a rectilinear motion) x  f (t ) , for which we can define the
velocity v 
d 2x
dx
and the acceleration a  2 . In a typical problem, we know the force F (t )  ma(t ) , so the
dt
dt
acceleration.
In particular, if the acceleration is constant, we find:
(8)
dv
dx
at 2
av
 at  C1  x 
 C1t  C2 .
dt
dt
2
If we consider the initial conditions x  0  x0 and v  t0   v0 , the solution becomes x 
at 2
 v0t  x0 .
2
Typical problems with constant acceleration are either rectilinear motions free fall, projectile motion, or other vertical
motion with constant acceleration in general.
In this case, (9) F  mg 
x  g
t2
 C1t  C2 .
2
We’ll use g  32 ft / sec2 or g  9.8m / sec2 .
Example: Problem 24 / Page 17
x0  400 ft, v0  0, T  ? s.t. x(T )  0, v(T )  speed (T )  ?
Using (9), we find x  16t 2  400  16  t 2  25  (in feet)
x  0 for T=  5 and 5, and therefore choose T  5 . v(t )  32t  v(5)  160  s(5)  v(5)  160 .
Homework: Problems 2,4,6,8,10,11,16,20,26 and 32 from Section 1.2 in the Textbook.
9
1.3 Slope Fields and Solution curves of y '  f ( x, y )
Some first order de’s (1) y '  f ( x, y ) may be very hard or even impossible to integrate (solve) at least by the methods
we’ll learn in this chapter. In any event, the graphical method of slope (direction) fields that we will learn in this
section is a welcomed aid or tool to obtain a qualitative picture of the solution.
Since y '( x)  f ( x, y ) represents the slope of the tangent line to G f at x , by evaluating f ( x, y ) (say at a sequence of
points in a rectangular grid on the xOy axis), we’ll obtain the slopes of the tangent lines of the solution at those points
and therefore the behavior of the solution. For this to work, we usually need a large enough number of points, and
therefore either done by hand at enough points or by computer. Note that this can be done readily for any equation of
type (1) without any knowledge of the solution y ( x) .
Example:
Slope fields for autonomous equations y '  f ( y ) are particularly simple and revealing.
Examples: (let students do)
Draw a slope field and representative solution curves for y '   y , y '  2 y , y '  1  y and y '  y ( y  3) .
These slope fields and solutions apply to population models and other models.
Using a software, slope fields can be plotted in Mathematica (you will need to learn this), with the commands (write
exactly as they are):
<<Graphics`PlotField`
VectorPlot [{1,x*y},{x,-2.5,2.5},{y,-2.5,2.5},PlotPoints->20,Axes->True,Frame->True,ColorFunction->Hue]
(let them try Mathematica for all of the above) and maybe for y '  sin( x) sin( y ) .
Also, you can use the applet at math.rice.edu/~dfield/dfpp.html go to dfield , press ok, input y’, x ind variable, ranges.
Clicking anywhere in the Slope Field shows you the solution curve that goes through that point.
Therefore, get proficient with Mathematica’s VectorPlot (I do not recommend calculator), you can also check your
work online.
However, if we want to still graph a direction field by hand, a more efficient method to graph it is by using isoclines.
For the general equation (1), isoclines are curves in the xOy plane where f ( x, y )  C  constant , so we are plotting
curves with a same constant slope for y ( x) , in order.
10
Example:
3
y'  x y
2
x-y=0  y=x slope is 0
1
x-y=1  y=x-1 slope is 1
x-y=2  y=x-2 slope is 2
0
x-y=-1  y=x+1 slope is -1
x-y=-2  y=x+2 slope is -2
1
2
3
3
2
1
0
1
2
3
Figure 1.3.2: Slope Field for y '  x  y (Mathematica)
Calculate the direction field (by hand) and obtain a solution curve for y '  xy .
x
-2
-2
-2
-2
-2
-1
-1
-1
-1
-1
0
0
0
0
y
-2
-1
0
1
2
-2
-1
0
1
2
-2
-1
0
1
y’
4
2
0
-2
-4
2
1
0
-1
-2
0
0
0
0
x
0
1
1
1
1
1
2
2
2
2
2
y
2
-2
-1
0
1
2
-2
-1
0
1
2
y’
0
-2
-1
0
1
2
-4
-2
0
2
4
2
1
0
1
2
2
1
0
1
2
Figure 1.3.1: Slope field for y '  xy (Mathematica)
11
You should know how to plot slope fields by hand, especially using isoclines, and what they mean, also to plot
corresponding solution curves. You should know how to use Mathematica and online applet to check your work.
Note that slope fields give only a qualitative picture of the solution, which may sometimes be misleading for more
complicated (i.e. non-linear equations). Remember that even if a function f ( x, y ) exists (and even if it is continuous)
(e.g. we can construct a slope field) , the solution may not necessarily exist everywhere. Therefore, a slope field gives
a qualitative picture of a solution, but for a precise and accurate graph of the solution, we will need to solve the
1
equation (if possible). Example: y '  y 2 has a nice slope field, but the solution is infinite (discontinuous at
).
y0
Homework: Problems 1,2,5,9,10,21,22 (isoclines for these two),23 and 26. Also fill in the steps of investigation A on
Page 30.
12
1.4 Linear first order equations y ' p( x) y( x)  q( x) and applications
In this section we will learn:
A. A specific method to solve linear first order equations of the form (1) y ' p( x) y ( x)  q( x)
B. Examples
C. Existence and Uniqueness theorem for (1) (optional)
D. Applications (mainly mixture problems)
A. In this section we learn how to solve the linear first order equation: (1) y ' p( x) y  q( x) .
Since the left hand side looks like the derivative of a product, we try to obtain that product. It looks like  p( x) y  ' , but
it doesn’t work right away. We need to multiply first (1) by
 ( x)  e 
p ( x ) dx
, which is called an integrating factor of
the de (1), since after (1) is multiplied by  ( x ) , it becomes readily integrable.
p ( x ) dx
p ( x ) dx
p ( x ) dx
After multiplying (1) by  ( x ) , it becomes e
y ' e
p ( x) y  e 
q( x) which can be written as
p ( x ) dx
p ( x ) dx
d   p ( x ) dx   p ( x ) dx
y  q( x)e
dx  C
y  e
q( x) which can be integrated immediately to produce e
e
dx 


  p ( x ) dx 
 p ( x ) dx dx  C 
so (2) y  e
 q( x)e
.


Note that the constant of integration present in  ( x ) does not have any effect (it means multiplying with an extra
constant), so we don’t write it. The other constant of integration is important, so we do write it.
This is the final form of the solution of (1), which has one constant of integration C. Do NOT memorize (2), but just
the method to obtain it. Remember that in order to obtain the correct integrating factor, we need to have exactly the
form of the equation (1) to start it, if your equation is not in that form, first bring it to form (1).
Note that the method needs to calculate two integrals which we assume to be possible, if not, we need to resort to
table of functions or to numerical methods, in both of these methods we replace the integrals with
t
 p ( x ) dx dx  q(t )e 0
q
(
x
)
e

 p( x)dx   p(t )dt  C and 
x
0
x
p ( z ) dz
dt  C .
0
B. Examples:
13
1. Solve the differential equation: (3)
dy 1
1
 y  e x /3 . Plot several solutions and choose the particular solution
dx 2
2
whose graph contains the point (0,1).
1
dx
d x /2
1
1
Solution: Equation (3) is already of form (1), so we multiply by e  2  e x /2 to obtain:
e y   e x /2 x /3  e5 x /6

dx
2
2
1 6 5 x /6
3
3
3
 e x /2 y 
e  C  e5 x /6  C  y  e5 x /6 x /2  Ce  x /2  (4) y  e x /3  Ce  x /2
2 5
5
5
5
It is a good idea (if have time) at end to verify that this solution is correct, in other words, verify that (4) verifies (3).
(let them do that).
A graph of direction field, typical solution and desired solution are shown below (matlab generated with dfield7).
Figure 1.4.1: Matlab generated direction field and solution curves for (3)
The solution through (0,1) is in blue (thin line).
2. Find a general solution of the differential equation (5)  x 2  1
dy
 3 xy  6 x .
dx
After bringing to form (1) and performing same steps we find y  2  C  x 2  1
3/2
.
Direction field and typical solutions are shown below:
Figure 1.4.2: Matlab generated direction field and solution
curves
14
for (5) .
C. Existence and uniqueness theorem (optional now):
We have discussed generally that some ivp may not have a solution at all or even if they have a solution, they may not
have an unique solution. For general first order linear ode’s of the form (1), this theorem is particularly simple:
Theorem 1.4.1 (existence and uniqueness for linear first order equations):
If the functions p ( x) and q ( x) are continuous on an open interval I that contains the point x0 , then the initial value
problem (6) y ' p( x) y  q( x) , y( x0 )  y0 has a unique solution y ( x) on I, given by the formula (2) with the
appropriate value of C.
Proof: The proof is given by the method to find the solution (so we see that y ( x) in (2) verifies the equation), and,
under the conditions of the theorem, y ( x) exists on I and we can find C such that y(x) verifies the initial condition in
(6) (in general then y  Q( x)  C P ( x) , where Q(x) and P(x) are continuous at x0 . Then we can satisfy y( x0 )  y0 for
ANY  x0 , y0  ).
Examples:
1. Use Theorem 1.4.1 to find an interval in which the initial value problem xy ' 2 y  4 x 2 , y (1)  2 has a unique
solution. Then verify this claim by finding the solution. Then repeat the procedure for the initial condition y (1)  2 .
2. Investigate the existence and uniqueness of solutions of the first order de xy ' y  1 .
According to Theorem 1.4.1, we are guaranteed a unique solution on any interval that contains x0 and does not
contain 0. This is possible as long as x0  0 . When x0  0 , a solution (or a unique solution) is not guaranteed.
This is apparent in the direction field below. The solution appear not to be defined at x=0 (not surprisingly, think what
the slope is there).
Solve the equation to elucidate this and consider
y( x0 )  y0 for x0  0 and x0  0 successively.
4
Note that for (1) p(x) and q(x) are functions of x only, so
we may have a restriction of the interval on the x axis only.
2
0
2
4
4
2
0
2
4
15
Figure 1.4.3: Direction field for xy ' y  1 (Mathematica)
D. Applications: (mixture problems):
One important application of linear first order equations of form (1) is mixture problems (which can .
Assume that at time t=0 a tank contains Q0 lb of salt dissolved in V0 gal of water. Assume that brine containing cin lbs
of salt/gal is entering the tank at a rate of rin gal/min and that the well stirred mixture is draining from the tank at the
same rate. Set up the ivp that describes this flow process. Find the amount of salt Q(t) at any time. (in particular
problems we may also be interested in the amount of salt presented in the tank after a very long time).
Note that there are two quantities possibly varying in this problem: the amount of water (in the tank) and the amount
of salt (in the water).
When setting up de’s for these quantities, it is helpful to match the units of measure. This is true for any other
problems.
In our case:
dV
 rin  rout
dt
 gal / min  =0 (since
dQ
Q(t )
 cin rin  cout rout  cin rin 
rout
dt
V0
rout  rin )  V (t )  V0
lbs / min  (which is a first order linear equation of form (1)) Q(t )
In the more general case, when rout  rin :
dV
 rin  rout
dt
 gal / min   0  V (t ) (using V0 )
dQ
Q(t )
 cin rin  cout rout  cin rin 
rout
dt
V (t )
lbs / min  .
Examples:
1. Example 5/52:
V0  90 gal , Q0  90lbs , cin  2lb / gal , rin  4 gal / min , rout  3gal / min Q(T)=? For V (T )  120 .
Homework: Problems 1,4,6,11,16,17,25,30,33,34,36,38,43,44 from Section 1.5/Page 54
16
1.5 Separable equations 𝑦 ′ = 𝑃(𝑥) ∗ 𝑄(𝑦)
In this section we will learn:
A. Method
B. Examples
C. Applications
A. In the previous sections we have solved direct integration equations y '  f ( x) and linear first order equations
y '  p( x) y ( x)  q( x) . In this section we are looking at another particular first order equation: separable equations.
Separable equations are equations of the form: (1) y '  f ( x) g ( y ) .
In order to solve (1), we write it as
dy
dy
 f ( x) g ( y ) and fully separate the variables to get
 f ( x)dx
dx
g ( y)
dy
 f ( x)dx  (2) G ( y )  F ( x)  C , so in principle, we have a very simple method to solve (1), as long as
g ( y) 
we recognize it as such and are able to calculate the integrals.

Remarks:
 Separable equations are often non-linear and therefore their solutions will often have a more interesting
behavior (develop singularities or may cross one another)
 Equation (2) is implicit and often we cannot transform it to an explicit equation y  y ( x ) easily
 We usually loose the particular solution g ( y )  0 when dividing by g ( y ) in the first step. It is good to keep
this in mind and to add these solutions at the end. (Think of this in analogy with an algebraic equation, e.g.
y  sin( y) 1  0 ).
B. Examples:
1. y '  y  1 
 y( x)  1  Ce x the solution y  1 is not lost here (we can have C=0)
2. y '  6 xy , y (0)  7 . We note that this equation is also linear, but we solve it as separable:
… We get the general solution y( x)  Ce3 x and after a general i.c. y(0)  y0 : y( x)  y0e3 x so in our case the
2
2
particular solution is y( x)  7e3x . Graph solutions in a direction field and choose the particular solution.
2
dy x 2  1
1
 2 , y(1)=2   y 3  y  x   C . This solution is implicit for y(x) and cannot be transformed
x
dx 3 y  1
easily into an explicit solution. However we can use the i.c. and find C=10. Note that the solution has a discontinuity
at x  0 . This is because f ( x, y ) is discontinuous at x=0 (later). We did not loose any particular solution here by
division. Graph solutions in a direction field and graph the particular solution (Maple …)
3. x 2
17
2
 x2

dy
y
(
x
)

1

4.
 2 x y  1 (we will hopefully need – find y  1 ). Integrating, we find
  C  . Note that the
dx
 2

particular solution y  1 has been lost here by division (there is no C for which this would exist).
C. Applications:
Many practical problems produce separable differential equations. An important of separable de’s commonly found
in applications are the autonomous equations y '  f ( y ) , and these will be our focus here.
C1. Populations models: Growth and decay:
If we assume constant birth and death rates for a given population, then over a short period we find:
P       Pt , where  and  are the birth and death rates, respectively per individual per unit time.
dP
 kP , where k     , which has the solution P  P0 e kt . k here represents the rate of
dt
growth (decay if <0) of a particular population. If k >0 we can define a doubling time which depends on k only.
Note that the doubling time remains the same as we advance in time (for any starting time).
This produces the de: (2)
C2. Radioactive decay:
dN
 kN ,
dt
where again k >0. The solution of (3) is N  N 0 e  kt . k is the rate of decay here and it affects a halving time for the
Similarly, the amount of a radioactive substance can decay proportionally to the current amount: (3)
substance, often called half-life, again the same for any starting time.
C3. Other population models: (logistic model):
This improves C1 by imposing a cap on the total population (due to predators, limiting in resources, etc). The
equation governing this model is y '  ky  M  y  . Solve this yourself, find y(x) and plot.
C4. Newton’s law of cooling:
The temperature of a certain object brought into an environment with a different temperature A evolves according to
dT
 k  A  T  , where k  0 .
the differential equation:
dt
Do a direction field (by hand) and solve to find T (t )  A  Ce kt . Here C  A  T0 . Plot.
Example: Problem 7/39. A  375 (C ) , T (75)  125 , T (0)  50o . T (?)  150o .
Solve the Newton’s law of cooling equation to find T  375  325e  kt , using A and T0 . Then using T (75)  125
produces k  
1  10 
ln    0.0035  0 , so T  375  325e0.0035t . Solving
75  13 
T (t )  150 produces e  kt 
43
1  43 
so t   ln   105 min . So total mins after 5:00 pm are 105 min so about 6:45 pm.
65
k  65 
Homework: Problems 1,2,4,5,12,15,20,22,27,29,32,33,36,44,49,54
18
1.6 Exact equations
In this section we will learn:
A. A specific method to solve exact equations of the form: (1) M ( x, y )dx  N ( x, y )dy  0
B. Examples
A. Exact differential equations are equations of form (1) with an extra condition. To try to integrate (1), write (1) as
dy
 0 . As usual y  y ( x ) is the unknown function which needs to be determined. Also note that
dx
equation (2) can be constructed from any general equation y '  f ( x, y ) .
(2) M ( x, y )  N ( x, y )
We are searching for a solution of (2) in the special case in which (2) represents (3)
(4)
dF ( x, y )
 0 . Remember that
dx
dF ( x, y ( x)) F ( x, y ) F ( x, y ) dy
, according to the chain rule for a function of two variables. If (2) is of the


dx
x
y
dx
F ( x, y )
F ( x, y )
. In this case, the equation (2) is called exact and it
and N ( x, y) 
x
y
has the solution (6) F ( x, y )  cons tan t .
form (3)-(4), then (5) M ( x, y ) 
Question: How can we determine in advance if an equation of form (2) if it is exact, that is if a function F ( x, y ) exists
such that (5) hold?
We start by posing the converse problem: If (5) hold and
the mixed partial derivatives
F ( x, y )
F ( x, y )
are continuous, then it is known that
and
x
y
 2 F ( x, y) M ( x, y)
 2 F ( x, y) N ( x, y)
need be equal. Therefore, a necessary

and

yx
y
xy
x
M ( x, y ) N ( x, y )
. It turns out that (7) is also sufficient (see proof in the

y
x
textbook, based on construction), in other words, if (7) holds then we can find a function F ( x, y ) such that (5) hold,
and therefore the solution of (2) is (6).
condition for (2) to be exact is (7)
Let’s give the main theorem and a few examples:
Theorem 1.6.1:
Let M ( x, y ), N ( x, y ), M y ( x, y) 
M ( x, y )
N ( x, y )
and N x ( x, y) 
be continous on a region R  [a, b]  [c, d ] .
y
x
Then the de (2) is exact on R (i.e. there exists a function F ( x, y ) such that (2) is equivalent with (3)) if and only if
19
(7)
M ( x, y ) N ( x, y )
hold.

y
x
Remark: When solving an equation (1) by this method, two steps are necessary:
 First, verify that (1) is exact, that is verify that (7) holds.
 If (7) holds, the equation is exact, and we find the function F ( x, y ) by solving (5).
 If (7) does not hold, we can try to transform (1) to an exact equation by multiplying with an integration factor
(about this a bit later).
B. Examples:
1. Solve the differential equation: (8)  6 xy  y 3  dx   4 y  3x 2  3xy 2  dy  0 .
In this case, M ( x, y)  6 xy  y 3 and N ( x, y)  4 y  3x 2  3xy 2 . In order to see if (5) are true for some F ( x, y ) , verify
condition (7).
In this case, M y ( x, y)  6 x  3 y 2 and N x ( x, y)  6 x  3 y 2 , therefore M y  N x and (8) is exact. We use (5) to find
F ( x, y )
=M ( x, y )  6 xy  y 3 . Integrate this with respect to x to find
x
F ( x, y )
F ( x, y)  3x 2 y  xy 3  g ( y) . Imposing now (5) 2 :
=N( x, y )  4 y  3x 2  3xy 2 we find that
y
F ( x, y ) . According to (5)1 :
3x 2  3xy 2  g '( y)  4 y  3x 2  3xy 2  g '( y)  4 y  g ( y)  2 y 2  const . Therefore
(9) F ( x, y)  3x 2 y  xy3  2 y 2  const and the solution of (8) is given by (6) with F ( x, y ) given by (9), which is the
implicit equation: 3x 2 y  xy 3  2 y 2  const .
Can draw a direction field and solution curves (note non-uniqueness in a point).
2. Solve (10) y3dx  3xy 2 dy  0 . In this case M ( x, y)  y3 so M y  3 y 2 and N ( x, y )  3xy 2 so N x  3 y 2 so M y  N x
and (10) is exact . .. Here F ( x, y)  xy 3  const , so a solution of (10) is xy3  const .
Note however that (10) can naturally be divided by y 2 to become ydx  3xdy  0 which is not exact anymore ( 1  3 ).
The idea is that there are equations of form (1) which do not appear exact but which can be made exact by
multiplication with an integrating factor.
Extra credit: Analyze the integrating factor problem in general, that is consider an equation (1) which is NOT exact
M y  N x , multiply (1) by  ( x, y ) and find a de that  ( x, y ) has to satisfy in order for (1) to be exact. Find a solution
(method to solve of) this equation only in the two special cases    ( x) and    ( y ) . (it turns out
M  Nx
d
d  M y  Nx

 and
 y
 . In particular, find an appropriate integrating factor and solve:
dx
N
dy
M
20
(a) x 2 y 3  x 1  y 2  y '  0 and (b)  x  2 sin ydx  x cos ydy  0 . If time, cover 33 and/or 36.
Homework: Problems 31,32,(33 or 36) 35,38, 42 from Section 1.6 / Page 70.
1.7 Substitution methods
Until now, we have learnt a few special methods to solve a few types of (mostly) first order ode’s, namely: direct
integration, linear first order ode’s, separable equations and exact equations.
In this section, we will learn a few other methods which make use of substitution to bring our equation to one of these
already learnt methods. Note that the methods presented here are not exhaustive, other (substitution methods exist),
see problems in textbook for Ricatti or Clairaut. Historically, substitution methods are very important as they were the
first to be used to solve first order ode’s.
We divide our types of substitution as:
i) General Main Idea, simple obvious substitution;
ii) Homogeneous equations;
iii) Bernoulli equations;
iv) Reducible second order equations.
i) Simple Substitution:
For some de’s, there is an obvious direct substitution which brings the de to a known form.
Examples:
1. The de (1)
dy
2
  x  y  3 is NOT of the learnt types (not direct integration, not linear, not separable and –perhaps
dx
– not exact).
We try v( x)  x  y  3 (instead of y ( x) ), then
dv
dy
dy dv
 1


 1 , so that (1) becomes
dx
dx
dx dx
dv
 1  v 2  v '  1  v 2 which is an (autonomous) separable equations and produces:
dx
dv
 dx  arctan  v   x  C  v  tan  x  C   y  tan  x  C   x  3
1  v2
2. More generally, any first order ode of the form (2)
dy
 F  ax  by  c  becomes autonomous (separable) after the
dx
substitution v  ax  by  c .
3. Sometimes a direct substitution is a bit less obvious:
21
30/1.6: (3)  x  e y  y '  xe  y  1 . Again, none of the learnt forms.
Since y appears only through e y , we try v  e y  v '  e y y '  y ' 
 x  v
v' v'
 . Substituting this in (3) we find:
ey v
v' x
x  x
  1  v '   1   1 , then either v   x  y  ln   x   for x  0 -- particular solution, or
v v
v  v
v '  1  v  x  C  y  ln  x  C  , for x  C  0 -- general solution. (need v  x  C  0 by definition of v ).
You can verify that both the particular (singular solution here) and the general solution are solutions of (3).
Look also at Example 6.
ii) Homogeneous equations:
 y
A first order ode is called homogeneous if it is of the form: (4) y '  F   .
x
Note that some equations do not appear in this form and need to be brought to form (4).
We solve (4) by the substitution v 
y
xy ' y y ' y
  2  xv '  y ' v  y '  v  xv ' .
. Doing this, we find v ' 
x
x2
x x
Substituting this in (4) we find v  xv '  F (v)  xv '  F (v)  v , which is a separable equation.
Examples:
1. (Example 2) 2 xyy '  4 x 2  3 y 2 .
Note that this does not appear homogeneous, but after division by 2 xy : y '  2
this first. We do v 
y

x
  y '  v  xv '  v  xv ' 
x 3y
(homogeneous) – always verify

y 2x
2 3
 v
v 2
2 v 4  v2
2v
dx
 xv '   

dv 
 ln  4  v 2   ln x  C
2
v 2
2v
4v
x
 4  v 2  Kx  K  0   v   Kx  4  y   x Kx  4
2. xy '  y  x 2  y 2 , for y  x0   y0 , x0  0 .
22
Dividing by x : y ' 

dv
1 v
2

y
y2
y
 1  2 and doing v   v  xv '  v  1  v2  xv '  1  v2 (separable)
x
x
x
dx
 arcsin(v)  ln x  C  v  sin  ln x  C   y  x sin  ln x  C  . For
x
x0  0  x  0  y  x sin  ln( x)  C  . If y  x0   0 , find C   ln( x0 ) .
iii) Bernoulli equations:
Bernoulli equations are equations of the form (5) y ' P( x) y  Q( x) y n
(they look like linear only that they have Q( x) y n instead of Q( x) y ), where n is a real number (usually rational).
In this case, the correct substitution is not v  y n but (6) v  y1n . Doing (6), we have
v '  1  n  y  n y '  y ' 
1 n
1 n
y v ' , so (5) becomes
y v ' P( x) y  Q( x) y n , dividing by y n  0 we obtain:
1 n
1 n
(7) v ' (1  n) P( x)v  (1  n)Q( x) .
(7) is the linear first order ode produced by (5) after substitution (6).
Example:
1. 20/1.6
(8) y 2 y ' 2 xy 3  6 x . This doesn’t look like (5) yet, but after division by y 2 we get (9) y ' 2 xy  6 xy 2 which is a
Bernoulli equation with n=-2. Therefore, we use the substitution v  y1 n  y 3  v '  3 y 2 y '  y ' 
1 2
y v ' so (9)
3
1
becomes y 2 v ' 2 xy  6 xy 2 , multiply by 3y 2 to get (10) v ' 6 xv  18 x , which is a linear first order ode.
3
3
Multiplying by i.f. e3x gives v  3  Ce 3 x  y  3  Ce 3 x
2
2
2
6
4/3
4/3
1/3
2. Example 5 (book): xy ' 6 y  3xy  y ' x y  3 y (Bernoulli with n  4 / 3 ) then do v  y to get
1
v '   y 4/3 y '  y '  3 y 4/3v ' and the equation becomes …
3
2
1
1
v ' v  1   v  x  Cx 2  y1/3 
y
2
3
x
x  Cx
x  Cx 2
______________________________________________________________________________________________
As a general conclusion in this section, note that it is very important to identify first if our equation is direct
substitution, homogeneous or Bernoulli or some other type in order to know which method to use. Some de’s are
simultaneously of more than one type, so they can be solved by multiple methods (see e.g. Example 4 in book, which
is simultaneously homogeneous and Bernoulli with n=-1). Also, look at end of section (before starting homework) to
decide what type of equation each of the de’s from problems 1 to 30 are.
23
Our last type of equation:
iv) Reducible second order equations:
These are special second order equations (11) F ( x, y, y ', y '')  0 which can be reduced to first order by substitution.
We look at two main cases:
a) Dependent variable y missing: Then (11) is of the form: (12) F ( x, y ', y '')  0 .
We use the substitution p  y ' to bring (12) to the first order ode (13) F ( x, p, p ')  0 .
Example: 47/1.6
y ''   y ' . Do p  y ' to get p '  p 2 and then by separation of variables:
2
dp
1
1
1
 dx    x  C1  p  
 y' 
 y   ln x  C1  C2
2
p
p
x  C1
x  C1
b) Independent variable x missing: Then (11) is of the form: (14) F ( y, y ', y '')  0 .
We do the substitution p 
dy
dp dp dy
dp
 y '' 

 p and view (14) with y as independent variable and
dx
dx dy dx
dy

dp 
p  p ( y ) as dependent variable, written in the form F  y, p, p   0 .
dy 

Example:
44/1.6
(15) yy ''  y '  0 . Do p 
2
py
dp
dy
 0  y  cons tan t (check (15) to see that this is indeed a solution)
 p 2  0 . So either p  0 
dx
dy
or y

dy
dp dp
dp
, then (15) becomes:
 y '' 

y'  p
dx
dx dy
dy
C
dp
dp
dy
 p 
   ln p   ln y  A  p  1
dy
p
y
y
dy C1
y2
  ydy  C1dx 
 C1 x  C2  y 2  C1 x  C2  y   C1x  C2 .
dx y
2
Therefore, in this section we have learnt direct substitution methods by:
-- direct substitution;
-- homogeneous equations;
24
-- Bernoulli equations;
-- reduction of order.
Also, in the previous sections we have learnt: direct integration, linear first order equations, separable equations and
exact equation.
Homework: Find ALL substitution types for problems 1-30 at the end of Section 1.6, verifying each of the learnt
types (your answer should be: equation (*) is e.g. Bernoulli and homogeneous but not reducible or direct integration).
Also, fully do problems 1,4,5,8,9,12,14,15,`6,19,24,25,28,29,43,45,48,51,54,55,57,59,63,64.
Prepare all material 1.1-1.7 in view of a Quiz/Test soon.
1.8 Existence and uniqueness for first order odes y '  f ( x, y )
We have learnt the existence and uniqueness theorem for linear first order ode’s. It says that if p ( x ) and q( x) are
continuous on I  x0 , then the initial value problem (1) y ' p( x) y  q( x), y ( x0 )  y0 has a unique solution on I .
In this section, we study a corresponding theorem for nonlinear equations.
Theorem 1.8.1 (existence and uniqueness for nonlinear first order ode’s):
Consider the initial value problem
(2) y '  f ( x, y), y  x0   y0 . If f  x, y  and f y  x, y  are continuous on some rectangle  x, y   R   a, b  c, d  , with
 x0 , y0   R , then there exists a h  0
such that (2) has a unique solution y   ( x) on  x0  h, x0  h  .
Remarks (differences between the linear and the nonlinear cases):
1. Note that in the nonlinear case (2) we are not guaranteed existence and uniqueness of solution on the entire R ,
as it was in the linear case (here the domain of existence and uniqueness of a solution is in general a restriction
of R ). It is difficult and often impossible to establish ahead of time, just based on f  x, y  , the restricted
domain, as this domain often depends on the initial condition as well. See for example y '  y 2 , y  0   y0 , for
 1
1
, therefore a restricted domain is  0,  for y0  0 .
y0
 y0 
2. This theorem reduces to the theorem for the linear case if (2) is linear, note that the necessary conditions for
that case are satisfied if p ( x ) and q( x) are continuous on I  x0 .
which R 

, but the solution does not exist at
However, in the nonlinear case, we can split these conditions clearer as:

f  x, y  continuous on R  existence  there is a h  0 such that (2) has a solution on  x0  h, x0  h  

f y  x, y  continuous on R  uniqueness  the solution above is unique on  x0  h, x0  h   .
3. As we have seen in the case of separable equations, nonlinear equations do not necessarily have all solutions
included in a general formula.
25
Remarks 1 and 2 above are particularly important and prominent in applications, that is solutions of non-linear
equations may develop singularities on R , and we can see that continuity of f  x, y  implies existence, while
continuity of f y  x, y  implies existence and uniqueness on a restricted domain. Let us look at a couple of examples.
Example 1: (nice solution, linear here so nonlinear theorem can be applied in the linear case):
(3) y '   y , the linear (or non-linear) theorem guarantees existence and uniqueness on
y  x0   y0 . This is because p ( x)  1 and q( x)  1 are continuous on
indeed a solution of on
(3) on
for any initial condition
. Solving the equation, we find  ( x)  Ce x ,
for any initial condition y  x0   y0 .
Example 2: (nice solution, nonlinear case):
(4) y ' 
3x 2  4 x  2
, y  0   1 .
2  y  1
Use the E&U theorem to predict if a (unique) solution exists on some rectangle around (0,-1).
(4) is a nonlinear equation. f  x, y  and f y  x, y  are continuous everywhere excepts at y  1 . Consequently,
according to Theorem 1.8.1, since a rectangle can be drawn around the initial point (0,-1) in which both f  x, y  and
f y  x, y  are continuous, then a unique solution exists in some rectangle around (0,-1). We can see that this is true.
(matlab direction field).
If solving this by hand, we get y 2  2 y  x3  2 x 2  2 x  C , from i.c. we get C=3  y 2  2 y  x3  2 x 2  2 x  3
  y  1  x3  2 x 2  2 x  4  y  1  x3  2 x 2  2 x  4 .
2
We can see that a unique solution exists in a rectangle around (0,-1). Only y  1  x3  2 x 2  2 x  4 satisfies the
given initial condition. To find where this solution is well defined find where the quantity under square root is
positive. The only real zero of this is when x=-2, so the desired interval is x>-2. At (-2,1) the tangent line to the graph
is vertical so at that point the solution is also not differentiable (infinite derivative). At that point the solution becomes
multi-valued.
Also note, that if the i.c is y (0)  1 instead, or any y  x0   1 , a solution would exist but it would not be unique in this
case (y=1) is exactly the value where the solution is not differentiable.
In conclusion, the theorem is valid in this example, although the interval is reduced (a unique solution is found on
 2,   -- interval of validity). A initial condition using the point of discontinuity of f  x, y  and
f y  x, y  gives a not
well defined solution (not unique).
26
Example 3: (5) y '  y1/3 , y (0)  0 .
Use the E&U theorem to predict if a (unique solution) exists on some rectangle around (0,0).
1 2/3
y , f  x, y  is continuous on  , but f y  x, y  is continuous everywhere
3
except on y  0 . Since the initial condition involves (0,0), the E&U theorem does guarantee a solution which passes
Since f  x, y   y1/3 and f y  x, y  
through (0,0), but not a unique solution at that point.
3
3
2

2 
Solving (5) in general, we find y    x  C  . Imposing y(0)=0 we find C=0  y    x  .
3

3 
Indeed, these two solutions together with the singular solution y  0 satisfy (5). Solutions of (5) with y  x0   0 exist
but are not unique, as the theorem predicts.
If we had taken y  x0   y0  0 as an initial condition, then we can find is some rectangle around  0, y0  in which a
unique solution of (5) exists.
Again the theorem is very helpful in providing points in the xOy plane where the existence (or uniqueness) of a
solution may fail.
(see also direction field for this equation).
y
Example 4: Look also at Example 6 in the book / 24: y '  2 , y ( x0 )  y0 . (E&U not guaranteed for x0  0 ).
x
We get y  Kx 2 , so the discussion is:
27
if x0  0 , then a unique solution is guaranteed in a rectangle around that point;
if x0  0 and y0  0 E&U are both not guaranteed and we actually see that no solution exists.
if x0  0 and y0  0 E&U are both not guaranteed, but multiple solutions (an infinity) exist.
Therefore, when starting with a non-linear de, verify the conditions of the theorem and use the theorem when
conditions are satisfied. When the conditions are not satisfied, we know that nothing is guaranteed by the theorem, but
all situations may occur, i.e. a solution may exist and be unique (still to find an example), a solution may exist and not
be unique (see above, often), or no solution exists (see last example).
Example 5: Can also give them y '  
y2
x
. x0  0  unique solution.
, y ( x0 )  y0 . We get y ( x)  
2
xC
x
x0  0 and y0  0  no solution.
x0  0 and y0  0  an infinity of solutions (for any C).
Homework: Problems 11, 12, 15, 17, 20, 32, 33 / page 27 and Problem 31/ Section1.4 / page 41 (Section 1.3)
and review Chapter 1 (except acceleration – velocity models for Chapter Test).
Optional: Proof of Theorem 1.8.1
The proof of the E&U theorem 1.8.1 is a good (first) topic for your project. Ask me for supplementary materials to
read for this proof if interested. This is a very important theorem and we should have an idea about the proof. (See
mostly Boyce and DePrima and maybe Coddington).
The lines of the proof are the following:
First, by a simple transformation, bring (2)  (6) y '  f ( x, y), y  0  0 .
Assume that the conditions of theorem 1.8.1 are met. We will construct a solution of the i.v.p. (2) in an iterative
fashion, by the method of Piccard’s iterations.
x
First (2) is equivalent with (6) y ( x)   f  s, y ( s )  ds . In order to find the solution of (6), construct a sequence
0
x
(7) yk 1 ( s)   f  s, yk ( s)  ds starting with y0 ( x)  0 (or any other). Note that all terms of the sequence satisfy the i.c.
0
We terminate the sequence when yk 1  yk in a convergence sense.
Questions:
 Do all members of (6) exist, or the process may break down at a point?
 Does the sequence in (6) converges?
 If it converges, is the limit function a solution of (2)?
28
 Is this limit solution unique? When .
The answers to these questions clarify the theorem.
 First, to calculate all terms of (7), we need f  s, yk (s)  to be well defined and continuous at all s  0, t  . Note
that f is guaranteed to be continuous on R=[a,b]x[c,d] at the beginning, but the danger is for some yk to get out
of R at some value of x inside [a,b] (to become infinite for a finite value of x in [a,b]). Then the process break
and this limits our domain (h appear). See also page and Boyce and DiPrima. Other conditions may insure that
this does not happen.
 The sequence can be showed to converge under the hypothesis (see Boyce and DiPrima).
 Clearly the solution is a solution of (2) since (6) implies (2) and the limit solution is a solution of (6)
 Uniqueness is guaranteed by continuity of f y  x, y  . That is why for existence (until now) we used and we
need only continuity of f  x, y  , but for uniqueness we need continuity of f y  x, y  . Continuity of f y  x, y 
implies (using a mean value theorem) that f  x, y  satisfies a Lipshitz condition on R, with M  max f y ( x, y )
(a continuous function on a bounded domain is bounded). Using Lipshitz, it’s easy to show uniqueness (see
Boyce and DiPrima).
1.9 Mathematical models: population models and velocity acceleration models
I. Population models:
We have looked at some population models in Section 1.5 (separable and autonomous equations).
Here we are going to briefly mention the different population models. We will look at three main population models:
1. Exponential growth:
If  (t ) and  (t ) are the birth and growth rate per individual per unit time, we find that the change in
population over a time interval t is P    (t )   (t )  Pt , which produces the differential equation:
dP
   (t )   (t )  P(t ) . For constant individual birth and growth rates we find the exponential growth
dt
model:
dP
 kP , where k     with solution P(t )  P0 e kt .
(2)
dt
(1)
More generally,  and  in (1) can still be functions of t or P(t).
2. Model (2) predicts exponential growth and decay, which we often find unsatisfactory (there is no limit in
population growth and other limiting factors such as finite resources are not considered).
As an extension, assume that  (t )  a  bP(t ) and  (t )   still constant. This means that the individual birth
rates decrease as the population increases. This (improved) model is called the logistic model, characterized by
the differential equation:
29
dP
 kP  M  P  . M here is the limiting population (carrying capacity). We will define it later as an
dt
equilibrium solution (start there, remain there) and stable.
Solve (3), graph slope fields and solution curves.
The table on page 76 gives a nice performance comparison of these two models.
3. Doomsday vs. extinction:
(3)
Assume that the rate of growth is proportional with number of encounters females males
k
P2
, and that the
2
death rate  is constant. We obtain a similar equation with (3) but with the signs reversed:
dP
(4)
 kP  P  M  called the doomsday vs. extinction model. This time M is an unstable equilibrium (
dt
P0  M means that the initial population is over a threshold where rate of growth overwhelms decay and
P0  M is the opposite situation).
Homework: Problems 1,4,7,9,11,16,22,26 on page 80 / Section 1.7
II. Velocity acceleration models
We will be looking at a couple of velocity - acceleration models. They are based on Newton’s second law of motion.
To present the ideas, consider a vertical motion in which the vector velocity is assumed to be oriented upwards (object
is moving upwards). We consider:
1. Only gravitational force: G  mgj . Since v  v j (assumed the downwards direction is negative), we find
dv
dv
  mg 
  g (here the velocity decreases as the object moves upwards since gravitation pulls it
dt
dt
down).
m
2. A little bit more sophisticated model considers also a friction force FR which always opposes the direction of
motion (oriented opposite to v ). It is usual to take FR  kv or FR  kv 2 .
For upward motion with FR  kv , we find:
dv
dv
k
dv
k
 mg  kv 
  g  v . Similarly, for FR  kv 2 , we find:
  g  v2 .
dt
dt
m
dt
m
dv
k
 g  v and similarly for FR  kv 2 :
(you can derive that the downward motion with FR  kv gives
dt
m
dv
k
 g  v 2 ).
dt
m
3. More generally, over long distances away from earth, the gravitational force cannot be assumed constant
mMG
anymore. In this case: G   2 j where M is the mass of the earth, m is the mass of the object and G is a
r
constant.
m
30
dv
MG
  2 (if the body would
dt
r
have moved toward earth it would accelerate due to gravitation, away from earth it decelerates). This equation
is niftily solved in the book.
For the upward motion (away from the gravitational field) we find: a  r '' 
Homework: Problems 1,3,7,19,20,27 and 30 as Extra Credit from Section 1.8 / page 90
31