Slope Fields & Euler's Method for Diff. Equations in Linear Algebra - Prof. Bryan J. Bornh, Study notes of Linear Algebra

The concept of slope fields and solutions curves for first-order differential equations in the context of linear algebra and differential equations. It explains how to use the differential equation itself to sketch the slope field and construct piecewise linear approximation solutions using euler's method. The document also includes examples and exercises.

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Pre 2010

Uploaded on 07/30/2009

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MATH 2250 Linear Algebra and Differential Equations
Chapter 1: First-Order Differential Equations
1.3 Slope Fields and Solutions Curves
Given a first-order differential equation
),( yxf
dx
dy
,
the function f(x,y) is a formula for the slope of a solution curve y at any point (x,y) in the domain
of f.
Depending on the nature of the function f(x,y), we may be able to solve for y explicitly as a
function of x, say y(x). In many cases, we may obtain an implicit relationship between y and x in
which we can only view y as a function of x on a restricted region of the xy plane.
This section addresses the use of the differential equation itself to sketch the slope field of a
solution curve.
The diagram at right is the
slope field for
yx
dx
dy
.
Notice that the slopes are zero
along the line y = x and, in fact,
are constant along ALL lines
with slope m = 1.
We may use the slope field to
sketch graph of a solution curve
passing through a given point
(x,y) by following a short line
segment at the point to its other
end and following a new line
segment whose slope corresponds
to the new point. In this manner,
we may construct a piecewise-
linear approximation solution.
Sketch solutions to
yx
dx
dy
given that (a) y(-2) = 1 and (b) y(−2) = −3.
Notice that any two distinct solutions will not intersect and that all solutions become asymptotic
to the solution y = x. (WHY?)
pf2

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MATH 2250 Linear Algebra and Differential Equations

Chapter 1: First-Order Differential Equations

1.3 Slope Fields and Solutions Curves

Given a first-order differential equation

f ( x , y ) dx dy

the function f ( x , y ) is a formula for the slope of a solution curve y at any point ( x , y ) in the domain

of f.

Depending on the nature of the function f ( x , y ), we may be able to solve for y explicitly as a

function of x , say y ( x ). In many cases, we may obtain an implicit relationship between y and x in

which we can only view y as a function of x on a restricted region of the xy plane.

This section addresses the use of the differential equation itself to sketch the slope field of a

solution curve.

The diagram at right is the

slope field for

x y dx dy

Notice that the slopes are zero

along the line y = x and, in fact,

are constant along ALL lines

with slope m = 1.

We may use the slope field to

sketch graph of a solution curve

passing through a given point

( x , y ) by following a short line

segment at the point to its other

end and following a new line

segment whose slope corresponds

to the new point. In this manner,

we may construct a piecewise-

linear approximation solution.

Sketch solutions to x y

dx dy

  given that (a) y (-2) = 1 and (b) y (−2) = −3.

Notice that any two distinct solutions will not intersect and that all solutions become asymptotic

to the solution y = x. ( WHY? )

MATH 2250 Linear Algebra and Differential Equations

Euler’s Method: Numerically , since

x y dx dy

 , starting with an initial point ( x 0 , y 0 ), we may

compute numerical approximations to points on a solution curve using increments in x of  x^ as

follows:

y y f x y x x x x y k k k k k k           

1 ( ,^ )

1

where

f ( x , y ) dx dy

This method of approximation is know as Euler’s method and is covered in detail in 2.4.

Example: Given the differential equation

x y dx dy

starting with an initial condition of y (-2) = 1 and  x^ ^0.^5 , calculate the first 3 resulting points

( x 1 , y 1 ),( x 2 , y 2 ),( x 3 , y 3 ) using the recursive formulas above.

Solution: We need to calculate the slope (derivative) at each point in order to compute the next

point in the approximation.

1 0 0 0 1 0           

y y f x y x x x x y

so we obtain the point (−1.5, −0.5). Go up to the slope field for this differential equation and

sketch or trace the line with slope f (−2,1) = −3 passing through (−2,1) and verify that it passes

through the point (−1.5, −0.5).

The next point obtained by this method is

2 1 1 1 2 1           

y y f x y x x x x y

which is the point (−1, −1). Again, trace the corresponding line segment with slope

f (−1.5, −0.5) = −1 passing through (−1.5, −0.5) and verify that it passes through the

point (−1, −1).

The next point is given by

3 2 2 2 3 2           

y y f x y x x x x y

which is (−0.5, −1). Since the slope at the previous point is zero, the new point has the same y

coordinate and the x coordinate is  x^ ^0.^5 less.