Chapters 1 and 2

1 Introduction

Reading assignment: Study Sections 1.1 and 1.2. (If you have time it might be

good to read Section 1.3 to get an understanding of modeling applications of

Ordinary Differential Equations)

An Ordinary Differential Equation (ODE) is a relation between an independent variable

x and and a dependent variable y (i.e., y = y(x) depends on x) and its derivatives y(j) =djy

dxj

for j = 1, · · · , n. So it is an equation that can be written in the form

F (x, y, y(1), · · · , y(n)) = 0, (1)

For example, in this chapter we will learn to solve equations like the following:

y′ − x cos(x2) = 0, y′ − sin(x+ y) = 0, y′ − 2xy − x = 0, and y′ =x− yx+ y

.

The main reason to study differential equations due to the many practical physical ap-

plications. Solving ODEs can be very difficult or even impossible. But in this class we will

focus on the solution of very simple problems in order to give students some idea of what is

involved in the more general case. The main tools needed by the students is a background

in college algebra and calculus (differentiation and integration).

The goal is to find all functions y(x) satisfying the equation (1). Unfortunately this

objective is beyond reach other than some very special cases. In this class we will eventually

study a few of these special cases.

Notation and Terminology: The following discussion may seem a bit over the top. The

main purpose is to introduce the standard notation and terminology used in talking about

ODEs. DO NOT BE OVERWHELMED they are just words. Also read the book

where more examples are also given. The main words to understand are written in italics

and underlined.

Definition 1.1. 1. The order of highest order derivative, n, is called the order of the equation.

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2. If we can solve for the highest order derivative term, then we say the equation can be

put in normal form:dny

dxn= f(x, y, y(1), · · · , y(n−1)).

3. An ODE is said to be Linear if it can be written in the form

an(x)dny

dxn+ an−1

dn−1y

dxn−1+ · · ·+ a1(x)

d1y

dx1+ a0(x)y(x) = g(x). (2)

(a) A linear equation is is said to be homogeneous if g(x) = 0. If g(x) 6= 0 then the

equation is called non-homogeneous.

(b) A solution y(x) may only exist on a certain domain called the interval of existence.

The interval of existence may be open, e.g., (a, b), closed, e.g., [a, b] or it may be

open on one end and closed on the other, e.g., (a, b]. It is possible that this interval

is all of (−∞,∞).

4. Sometimes it is possible that y = 0 is a solution. In this case we call this the

trivial solution.

5. If we have a solution given in the form y = y(x) then we say that y(x) is an explicit solution.

But very often it is either difficult or even impossible to obtain an explicit solution

even if we know that it exists. In this case we may still be able to find a so-called

implicit solution. We define an implicit solution as follows:

A relation G(x, y) = 0 is called an implicit solution of the ODE (1) if there exists

a function ϕ(x) (whether we can find it or not) so that G(x, ϕ(x)) = 0 and also

F (x, ϕ(x), ϕ(1)(x), · · · , ϕ(n)(x)) = 0

6. Generally speaking an nth order ODE has an n-parameter family of solutions. That

is to say that the solution depends on n arbitrary constants (constants of integra-

tion). Thus we would write an implicit solution as G(x, y, c1, c2, · · · , cn) = 0 where

c1, c2, · · · , cn are arbitrary parameters. If we can find an implicit or explicit solution

containing n arbitrary parameters then we call the solution the general solution. For

example, a general explicit solution of y(3) = 0 is y = c1 + c2x+ c3x2.

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7. The Initial Value Problem (IVP) is to find a unique solution of (1) satisfying the n

constraints (called initial conditions)

y(x0) = y0, y′(x0) = y1, · · · , y(n−1)(x0) = yn. (3)

Remark 1.1. 1. In order to solve the IVP we generally we somehow find a general solu-

tion which depends on n parameters c1, c2, · · · , cn. We then use the n constraints (3)

to evaluate the parameters cj, j = 1, · · · , n to obtain a unique solution.

2. In case F (x, y) in (1) does not depend explicitly on the independent variable then (1) is

called autonomous, i.e., y′ = F (y). In many ways, as you will see later on, autonomous

equations are much easier to study than the nonautonomous case.

Example 1.1. The general solution to the nonautonomous differential equation y′ = 3x2

is y(x) = x3 + c where c (just integrate both sides) is an arbitrary constant. The unique

solution satisfying the initial condition y(2) = −1 is y(x) = x3 − 9.

2 First Order Equations and IVPs

Reading assignment: Your need to study Chapter 2, Sections 2.1 through 2.5.

In this section we will consider first order equations and, for the most part, we are

interested in those equations that can be put in the normal form

y′ = f(x, y) (4)

and the associated IVP

y′ = f(x, y), y(x0) = y0. (5)

Theorem 2.1 (Fundamental Existence Uniqueness Theorem). Let R be a rectangular region

R = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d} that contains the point (x0, y0) in its interior. If f(x, y)

and ∂f/∂y are continuous in R, then there exists an interval I0 = (x0 − h, x + h) ⊂ (a, b)

for a number h > 0, and a unique function y(x) defined on I0 that solves the IVP (5) on I0.

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The first order autonomous equations y′ = f(y) are particularly interesting and the

behavior of solutions can be described rather nicely even without solving the equation.

A fixed point (or equilibrium point) of a differential equation y′ = f(y) is a root of the

equation f(y) = 0. Notice that an equilibrium point gives a solution to the differential

equation – it is a constant solution. In particular, suppose that y0 is an equilibrium point,

i.e., f(y0) = 0, then set y(x) = y0 for all x (a constant function). Since y(x) is a constant its

derivative is zero and y0 is an equilibrium point we have y′ = f(y). Qualitative information

about the equilibrium points of the differential equation y′ = f(y) can be obtained from

special diagrams called phase diagrams.

A phase line diagram for the autonomous equation y′ = f(y) is a line segment with labels

for so-called sinks, sources or nodes, one for each root of f(y) = 0, i.e. each equilibrium.

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1.4 Phase Line and Bifurcation Diagrams

Technical publications may use special diagrams to display qualitative

information about the equilibrium points of the di!erential equation

y! = f(y).(1)

This equation is independent of x, hence there are no external controlterms that depend on x. Due to the lack of external controls, the equa-tion is said to be self-governing or autonomous.

A phase line diagram for the autonomous equation y! = f(y) is a linesegment with labels sink, source or node, one for each root of f(y) = 0,i.e. each equilibrium; see Figure 11.

sinksource

y0 y1

Figure 11. A phase line diagram for an

autonomous equation y! = f(y).

The labels are borrowed from the theory of fluids, and they have thefollowing special definitions:6

Sink y = y0 The equilibrium y = y0 attracts nearby solutions at

x = !: for some H > 0, |y(0) " y0| < H implies

|y(x) " y0| decreases to 0 as x # !.

Source y = y1 The equilibrium y = y1 repels nearby solutions at

x = !: for some H > 0, |y(0) " y1| < H implies

that |y(x) " y1| increases as x # !.

Node y = y2 The equilibrium y = y2 is neither a sink nor a source.

In fluids, sink means fluid is lost and source means fluid is created. Amemory device for these concepts is the kitchen sink, wherein the faucetis the source and the drain is the sink. The stability test below inTheorem 2 is motivated by the vector calculus results Div(P) < 0 for asink and Div(P) > 0 for a source, where P is the velocity field of thefluid and Div is divergence.

Stability Test. The terms stable equilibrium and unstable equi-

librium refer to the predictable plots of nearby solutions. The termstable means that solutions that start near the equilibrium will staynearby as x # !. The term unstable means not stable. Therefore, asink is stable and a source is unstable.

Precisely, an equilibrium y0 is stable provided for given ! > 0 thereexists some H > 0 such that |y(0) " y0| < H implies y(x) exists forx $ 0 and |y(x) " y0| < !.

6In applied literature, the special monotonic behavior required in this text’s defi-nition of a sink is relaxed to limx!" |y(x) ! y0| = 0.

The simplest first order autonomous equation is when f(y) is a linear function of y. The

IVP can be written asdy

dx= ay, y(x0) = y0, a ∈ R.

The solution is given by

y(x) = y0eax.

This equation is a basic model for many important growth and decay problems (e.g., popu-

lation dynamics and the decay of radioactive materials). The main method to solve such a

problem is described in the next subsection.

In this section we briefly discuss the idea of using the notion of derivative of a function

as a tool to guess the shape of the solution to a first order equation

y′ = f(x, y).

Note that if ϕ is a solution of this equation then and ϕ(x0) = x0 then the slope of the tangent

line to the graph of y = ϕ(x) at (x, y) is given by f(x, y). Since we can compute f(x, y) at

every point we can plot the direction a solution will take starting from any given point. The

4

Direction Field for a first order ODE (4) is a figure in which arrows are placed at a grid of

points in the xy-plane with an arrow at each point (x, y) of the grid pointing in the direction

f(x, y). By starting at a point (x0, y0) one can move in the directions of the arrows to get

an idea what the solution of the IVP looks like.

Let us consider an example to demonstrate the idea.

y′ = y2 − x.

For this example we plot the solution satisfying y(0) = 0.

Notice that the above example is non-autonomous, i.e., f depends on x and y. As we

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have already mentioned, things are simpler for autonomous equations. Consider the example

y′ = sin(y).

We notice that the function f(y) = sin(y) which does explicitly depend on x. The

equilibria (or stationary) solutions are given by

sin(y) = 0 ⇒ y = kπ, k = ±1,±2, · · · .

For autonomous equations these solutions have very special properties. Generally, solu-

tions of the differential equation either approach of are repelled by the equilibrium solutions.

Direction Field for y′ = sin(y)

For this example we have plotted two solutions with initial conditions y(0) = 2 (Blue

curve) and y(0) = 5 (Green Curve), respectively. Notice that both solutions approach the

sink corresponding to the stationary solution y0 = π.

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It is sometimes useful to look at the graph of f(y) to determine the direction of the

arrows. What we gain from this picture is that if a solution begins in a region in which f(y)

is positive the y′ = f(y) > 0 so the function y(x) must be increasing. Notice that, other

than nodes, solutions are trapped between sources and sinks and these solutions move away

from sources and toward sinks.

++++++ ---

2.1 Separable Equations

A very important class of problem (autonomous and nonautonomous) are ones that can be

“separated.” These are problems that can be written in the form

dy

dx= f(y)g(x).

7

In this case we can rewrite the problem in the form

1

f(y)

dy

dx= g(x)

or1

f(y)

dy

dx= g(x).

Integrating both sides we arrive at

∫1

f(y)

dy

dxdx =

∫g(x) dx+ C

and we say the problem is solved “up to quadrature.” This simply means we must evaluate

the indefinite integrals.

This shows already how hard solving differential equations are since it is easy to write

down functions that cannot be integrated exactly in terms of elementary functions. For

example,

y = e−x2

has solution which can be expressed as

y(x) =

∫ x

0

e−s2

ds+ C

but it is well known that this integral cannot be expressed “in closed form.”

In spite of this example, the simplest class of separable first order equations are ones in

the form y′ = f(x) which can be written in the separated form dy = f(x) dx. Therefore by

the Fundamental Theorem of Calculus y =∫f(x) dx.

Example 2.1. Consider the differential equation y′ = xy2. We separate the dependent and

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independent variables and integrate to find the solution.

dy

dx= xy2

y−2 dy = x dx∫y−2 dy =

∫x dx+ c

−y−1 =x2

2+ c

y = − 1

x2/2 + c

Example 2.2. The equation y′ = y − y2 is separable.

y′

y − y2= 1

We expand in partial fractions and integrate.

(1

y− 1

y − 1

)y′ = 1

ln |y| − ln |y − 1| = x+ c

We have an implicit equation for y(x). Now we solve for y(x).

ln

∣∣∣∣ y

y − 1

∣∣∣∣ = x+ c∣∣∣∣ y

y − 1

∣∣∣∣ = εx+c

y

y − 1= ±εx+c

y

y − 1= cεx (Here we have substituted c for ±εc.)

y =cεx

cεx − 1

y =1

1 + cεx

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2.2 First Order Linear Equations

A first order, linear, homogeneous equation has the form

dy

dx+ p(x)y = 0.

Note that if we can find one solution, then any constant times that solution also satisfies the

equation. If fact, all the solutions of this equation differ only by multiplicative constants.

Notice that we can solve any equation of this type because it is separable.

dy

y= −p(x) dx

ln |y| = −∫p(x) dx+ c

y = ±e−∫p(x) dx+c

y = ce−∫p(x) dx

(Here we have renamed the arbitrary constant ±ec to c).

First Order, Linear Homogeneous Differential Equations. We have just shown that

the first order, linear, homogeneous differential equation,

dy

dx+ p(x)y = 0,

has the general solution

y = ce−∫p(x) dx. (6)

Notice that solutions differ by multiplicative constants.

Example 2.3. Consider the equation

dy

dx+

1

xy = 0.

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We use Equation (6) to determine the solution.

y(x) = c e−∫1/x dx, for x 6= 0

y(x) = c e− ln |x|

y(x) =c

|x|

y(x) =c

x

First Order, Linear Non-Homogeneous Differential Equations.

The first order, linear, nonhomogeneous differential equation has the form

dy

dx+ p(x)y = f(x). (7)

This equation is not separable. To solve Equation (7), we multiply by an integrating factor.

Multiplying a differential equation by its integrating factor has the effect of transforming

the left hand side of the equation into an exact derivative. By this we mean that multiplying

Equation (7) by the integrating factor, I(x), yields,

I(x)dy

dx+ p(x)I(x)y = f(x)I(x).

In order that I(x) be an integrating factor, it must satisfy

dy

dxI(x) = p(x)I(x).

This is a first order, linear, homogeneous equation with the solution

I(x) = cε∫p(x) dx.

This is an integrating factor for any constant c. For simplicity we will choose c = 1.

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So to solve the equation (7) we multiply by the integrating factor and integrate. Let

P (x) =∫p(x) dx.

eP (x) dy

dx+ p(x)eP (x)y = eP (x)f(x)

dy

dx

(eP (x)y

)= eP (x)f(x)

y = e−P (x)

∫eP (x)f(x) dx+ ce−P (x)

y ≡ yp + c yh

The general solution is the sum of a particular solution, yp, that satisfies y′ + p(x)y = f(x),

and an arbitrary constant times a homogeneous solution, yh, that satisfies y′ + p(x)y = 0.

Example 2.4. Consider the differential equation

y′ +1

xy = x2, x > 0.

First we find the integrating factor.

I(x) = exp

(∫1

xdx

)= εlnx = x

We multiply by the integrating factor and integrate.

dy

dx(xy) = x3

xy =1

4x4 + c

y =1

4x3 +

c

x.

The particular and homogeneous solutions are

yp =1

4x3 and yh =

1

x.

Note that the general solution to the differential equation is a one-parameter family of func-

tions.

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2.3 Exact Equations

Any first order ordinary differential equation of the first degree can be written (in infinitely

many ways) in differential form,

M(x, y) dx+N(x, y) dy = 0.

We know from Calculus III that if F (x, y) is a function satisfying

dF = M dx+N dy = 0,

then this equation is called exact. An (implicit) general solution of the differential equation

is

F (x, y) = c,

where c is an arbitrary constant. Since the differential of a function, F (x, y), is

dF ≡ ∂F

∂xdx+

∂F

∂ydy,

M and N are the partial derivatives of F :

M(x, y) =∂F

∂x, N(x, y) =

∂F

∂y.

Example 2.5.

xdx+ ydy = 0

is an exact differential equation since

d

dx

(1

2(x2 + y2)

)= x+ y

dy

dx

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The solution of the differential equation is

1

2(x2 + y2) = c.

A necessary and sufficient condition for exactness. Consider the exact equation,

M dx+Ndy = 0.

A necessary and sufficient condition for exactness is

∂M

∂y=∂N

∂x.

What this means is that if the equation is exact then we are guaranteed there exits a

function F (x, y) so that

M =∂F

∂xand N =

∂F

∂y(8)

and a general solution to the problem can be written as F (x, y) = 0.

More importantly (8) provides a method for finding F (x, y). Namely, integrating the

first equation of we see that

F (x, y) =

∫M(ξ, y) dξ + f(y),

for some f(y).

If we differentiate this equation with respect to y and use the the second equation in (8)

∂F

∂y= N(x, y)

we arrive at an equation

∫My(x, y) dx+ f ′(y) =

∂F

∂y= N(x, y).

It can be shown that, after simplifying, we will arrive at an equation only in y for which

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we can find f(y).

Let us reconsider the earlier example

Example 2.6.

xdx+ ydy = 0.

Here M = x and N = y so∂M

∂y= 0 =

∂N

∂x

so the equation is exact and we know there exists a function F (x, y) so that Fx = M and

Fy = N .

Integrating Fx = M with respect to x we have

F (x, y) =

∫x dx =

x2

2+ f(y).

Differentiating with respect to y and using Fy = N we have

y = N = Fy = f ′(y) ⇒ f ′(y) = y.

Thus, integrating we obtain

f(y) =y2

2

and we find

F (x, y) =x2

2+y2

2.

The implicit general solution isx2

2+y2

2= c

which, by renaming the constant c can be written as

x2 + y2 = c.

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2.4 Homogeneous Equations

Consider a first order equationdy

dx= f(x, y).

We say that this equation is Homogeneous if the right hand side can be written in the form

dy

dx= F (y/x). (9)

If this is possible the we can reduce the equation to a separable equation using the change

of dependent variable

u = y/x.

Note that u = y/x implies y = xu which, in turn, implies

dy

dx= u+ x

du

dx.

Therefore the equation (9) can be written as

u+ xdu

dx= F (u)

which is separable. Namely, we can write

du

F (u)− u=dx

x

and we obtain an implicit solution by integration

∫du

F (u)− u=

∫dx

x+ C.

Example 2.7. Consider the problem

xdy

dx=y2

x+ y.

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Dividing by x we obtaindy

dx=(yx

)2+(yx

).

Proceeding as above, we set u = y/x implies y = xu which, in turn, implies

dy

dx= u+ x

du

dx.

Thus the equation becomes

u+ xdu

dx= u2 + u.

Now we simplify by subtracting the u on both sides and separate the variables to obtain

du

u2=dx

x.

Integrating both sides we have

−1

u= ln |x|+ C.

So we get

u =−1

ln |x|+ C

and finally

y = xu =−x

ln |x|+ C

2.5 Bernoulli equations

A Bernoulli equation is any equation that can be written in the form

y′ + f(x)y = g(x)yα, α 6= 0, 1. (10)

Note that if α = 0, 1 then the equation is first order linear.

The Bernoulli equation can be reduced to a first order linear equation using the following

substitution:

u = y1−α.

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This relation impliesdu

dx= (1− α)y−α

dy

dx.

Multiplying the Bernoulli equation by (1− α)y−α we see that it can be written as

(1− α)y−αdy

dx+ (1− α)f(x)y1−α = (1− α)g(x)yαy−α

which simplifies todu

dx+ (1− α)f(x)u = (1− α)g(x).

If we let p(x) = (1− α)f(x) and q(x) = (1− α)g(x) then we obtain the first order linear

equationdu

dx+ p(x)u = q(x)

which can be solved using the methods discussed in the Section on first order linear problems.

Example 2.8. Consider the problem

y′ + y = y4.

This is a Bernoulli equation with α = 4 so (1 − α) = −3 and we set u = y−3. Then we

obtain the linear equationdu

dx− 3u = −3.

To solve this problem we find the integrating factor

I = e−3∫dx = e−3x.

Multiplying by I we obtain (e−3xu

)′= −3e−3x.

Next we integrate to obtain

∫ (e−3xu

)dx =

∫−3e−3x dx

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which gives

e−3xu = e−3x + C

or

u(x) = 1 + Ce3x.

Finally, converting back to y we have

y(x)−3 = 1 + Ce3x.

2.6 RHS Linear in x and y

If the equation (4) has the right hand side a linear function of in x and y, i.e., it has the form

F (ax+ by+ c) then a simple substitution transforms the problem into a separable problem.

Namely, if we set v = ax+ by + c then v′ = a+ bf(v) which is separable.

Example 2.9. Consider the problem

y′ = e−(x+y) − 1.

We set v = x+ y which implies v′ = 1 + y′ = 1 + (e−v − 1) = e−v. The equation v′ = e−v is

separable and seperating the variables we have

evdv

dx= 1 ⇒ ev dv = dx ⇒ ev = x+ c⇒ ex+y = x+ c.

Thus we have an implicit general solution ex+y = x+ c.

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