Math 308, Sections 301, 302,
Summer 2008
Lecture 2.
05/30/2008
Chapter 1. Introduction
Example 1.
(a) Determine for which values of m the function ϕ(x) = x m ,
where x 6= 0, is a solution to the equation
d 2y
dy
3x 2 2 + 11x
− 3y = 0.
dx
dx
(b) Show that exy + y = x − 1 is an implicit solution to
e−xy − y
dy
= −xy
.
dx
e
+x
Section 1.3 Direction Fields
One technique that is useful in graphing the solutions to a
first-order differential equation is to sketch the direction field for
the equation. To describe this method, we need to make a general
observation.
Namely, a first-order equation
dy
= f (x, y )
dx
specifies a slope at each point in the xy -plane where f is defined.
A plot of short line segments drawn at various points in the
xy -plane showing the slope of the solution curve there is called a
direction field for the differential equation. Because the direction
field gives the ”flow of solutions”, it facilitates the drawing of any
particular solution (such as the solution to an initial value
problem).
3
2
y(x)
1
K
3
K
2
K
0
1
1
2
x
K
1
K
2
K
3
Direction field for
dy
dx
= x2 − y
3
3
2
y(x)
1
K
3
K
2
K
0
1
1
2
x
K
1
K
2
K
3
Solutions to
dy
dx
= x2 − y
3
Method of Isoclines
Definition An isocline for the differential equation y ′ = f (x, y ) is
a set of points in the xy -plane where all the solutions have the
same slope dy /dx; thus, it is a level curve for the function f (x, y ).
For example, if y ′ = x + y , the isoclines are the straight lines
x + y = C , here C is an arbitrary constant. But C can be
interpreted as the numerical value of the slope dy /dx of every
solution curve as it crosses the isocline. To implement the method
of isoclines for sketching direction fields, we draw hash marks with
slope C along the isocline f (x, y ) = C for a few selected values of
C . If we when erase the underlying isocline curves, the hash marks
constitute a pard of the direction field for the differential equation.
3
2
y
1
K3
K2
K1
0
1
x
K1
K2
K3
Isoclines for y ′ = x + y
2
3
3
2
y
1
K3
K2
K1
0
1
x
2
K1
K2
K3
Direction field for y ′ = x + y
3
3
2
y(x)
1
K
3
K
2
K
0
1
1
2
x
K
1
K
2
K
3
Solutions to y ′ = x + y
3
Remark. When the isoclines curves are complicated, this method
is not practical.
Example 2. Draw the isoclines with their direction markers and
sketch several solution curves, including the the curve satisfying
the given initial condition y ′ = 2x 2 − y , y (0) = 0.
Section 1.4 The Approximation Method of Euler
Euler’s method (or the tangent line method) is a procedure for
constructing approximate solutions to an initial value problem for a
first-order differential equation
y ′ = f (x, y ),
y (x0 ) = y0 .
(1)
The main idea of this method is to construct a polygonal (broken
line) approximation to the solutions of the problem (1).
Assume that the the problem (1) has a unique solution ϕ(x) in
some interval centered at x0 . Let h be a fixed positive number
(called the step size) and consider the equally spaced points
xn := x0 + nh,
n = 0, 1, 2, . . .
The construction of values yn that approximate the solution values
ϕ(xn ) proceeds as follows. At the point (x0 , y0 ), the slope of the
solution to (1) is given by dy /dx = f (x0 , y0 ). Hence, the tangent
line to the curve y = ϕ(x) at the initial point (x0 , y0 ) is
y − y0 = f (x0 , y0 )(x − x0 ),
or
y = y0 + f (x0 , y0 )(x − x0 ).
Using the tangent line to approximate ϕ(x), we find that for the
point x1 = x0 + h
ϕ(x1 ) ≈ y1 := y0 + hf (x0 , y0 ).
Next, starting at the point (x1 , y1 ), we construct the line with
slope equal to f (x1 , y1 ). If we follow the line in stepping from x1 to
x2 = x1 + h, we arrive at the approximation
ϕ(x2 ) ≈ y2 := y1 + hf (x1 , y1 ).
Repeating the process, we get
ϕ(x3 ) ≈ y3 := y2 + hf (x2 , y2 ), etc.
This simple procedure is Euler’s method and can be summarized
by the recursive formulas
xn+1 := x0 + (n + 1)h,
yn+1 := yn + hf (xn , yn ),
n = 0, 1, 2, . . .
(2)
(3)
Polygonal-line approximation given by Euler’s method