Applications in Ch4

212 Chapter 4 Vector Spaces

4.8 Applications of Vector Spaces

Use the Wronskian to test a set of solutions of a linear

homogeneous differential equation for linear independence.

Identify and sketch the graph of a conic section and perform a

rotation of axes.

LINEAR DIFFERENTIAL EQUATIONS (CALCULUS)

A linear differential equation of order is of the form

where and are functions of with a common domain. If then the equation is homogeneous. Otherwise it is nonhomogeneous. A function iscalled a solution of the linear differential equation if the equation is satisfied when and its first derivatives are substituted into the equation.

A Second-Order Linear Differential Equation

Show that both and are solutions of the second-order linear differential equation

SOLUTION

For the function you have and So,

which means that is a solution of the differential equation. Similarly, foryou have

and

This implies that

So, is also a solution of the linear differential equation.

There are two important observations you can make about Example 1. The first isthat in the vector space of all twice differentiable functions defined on theentire real line, the two solutions and are linearly independent. Thismeans that the only solution of

that is valid for all is The second observation is that every linear combination of and is also a solution of the linear differential equation. To see this,let Then

Substituting into the differential equation produces

So, is a solution.The next theorem, which is stated without proof, generalizes these observations.y 5 C1e x 1 C2e2xy0 2 y 5 sC1e x 1 C2e2xd 2 sC1e x 1 C2e2xd 5 0.

y0 2 y 5 0

y0 5 C1e x 1 C2e2x. y9 5 C1e x 2 C2e2x y 5 C1ex 1 C2e2x

y 5 C1y1 1 C2y2.y2y1

C1 5 C2 5 0.x

C1y1 1 C2y2 5 0

y2 5 e2xy1 5 e xC0 s2`, `d

y2 5 e2xy20 2 y2 5 e2x 2 e2x 5 0.

y20 5 e2x.y29 5 2e2xy2 5 e2x,

y1 5 e xy10 2 y1 5 e x 2 e x 5 0

y10 5 e x.y19 5 e xy1 5 ex,

y0 2 y 5 0.y2 5 e2xy1 5 ex

n

yy

f sxd 5 0,xfg0, g1, . . . , gn21ysnd 1 gn21sxdysn21d 1 . . . 1 g1sxdy9 1 g0sxdy 5 fsxd

n

In light of the preceding theorem, you can see the importance of being able todetermine whether a set of solutions is linearly independent. Before describing a wayof testing for linear independence, you are given the following preliminary definition.

Finding the Wronskian of a Set of Functions

a. The Wronskian of the set is

b. The Wronskian of the set is

The Wronskian in part (a) of Example 2 is said to be identically equal to zero,because it is zero for any value of The Wronskian in part (b) is not identically equalto zero because values of exist for which this Wronskian is nonzero.

The next theorem shows how the Wronskian of a set of functions can be used totest for linear independence.

The proof of this theorem for the case where is left as an exercise. (See Exercise 40.)n 5 2

x

x.

W 5 |x10 x22x2 x33x26x| 5 2x 3.Hx, x 2, x 3J

W 5 |1 2 x210 1 1 x10 2 2 x210| 5 0.H1 2 x, 1 1 x, 2 2 xJ

4.8 Applications of Vector Spaces 213

REMARKThe Wronskian of a set of

functions is named after the

Polish mathematician Josef

Maria Wronski (17781853).

REMARKThis test does not apply to an

arbitrary set of functions. Each

of the functions

and must be a solution of

the same linear homogeneous

differential equation of order n.

yn

y1, y2, . . . ,

REMARKThe solution

is called the general solution of

the given differential equation.

C1y1 1 C2y2 1. . . 1 Cnyny 5

Solutions of a Linear Homogeneous Differential Equation

Every th-order linear homogeneous differential equation

has linearly independent solutions. Moreover, if is a set oflinearly independent solutions, then every solution is of the form

where are real numbers.CnC1, C2, . . . ,

y 5 C1y1 1 C2y2 1 . . . 1 Cnyn

Hy1, y2, . . . , ynJn

y snd 1 gn21sxdy sn21d 1 . . . 1 g1sxdy9 1 g0sxdy 5 0

n

Definition of the Wronskian of a Set of Functions

Let be a set of functions, each of which has derivativeson an interval The determinant

is called the Wronskian of the given set of functions.

Ws y1, y2, . . . , ynd 5 | y1y19...y1sn21d y2y29...y2sn21d . . .. . .. . . ynyn9...ynsn21d|I.

n 2 1Hy1, y2, . . . , ynJ

Wronskian Test for Linear Independence

Let be a set of solutions of an th-order linear homogeneousdifferential equation. This set is linearly independent if and only if theWronskian is not identically equal to zero.

nnHy1, y2, . . . , ynJ

Testing a Set of Solutions for Linear Independence

Determine whether is a set of linearly independent solutions of the linear homogeneous differential equation

SOLUTION

Begin by observing that each of the functions is a solution of (Try checking this.) Next, testing for linear independence produces the Wronskian of thethree functions, as follows.

Because is not identically equal to zero, the set

is linearly independent. Moreover, because this set consists of three linearly independent solutions of a third-order linear homogeneous differential equation, thegeneral solution is

where and are real numbers.

Testing a Set of Solutions for Linear Independence

Determine whether is a set of linearly independent solutions of thelinear homogeneous differential equation

SOLUTION

As in Example 3, begin by verifying that each of the functions is actually a solution of (This verification is left to you.) Testing for linear independence produces the Wronskian of the three functions, as follows.

So, the set is linearly dependent.

In Example 4, the Wronskian is used to determine that the set

is linearly dependent. Another way to determine the linear dependence of this set is toobserve that the third function is a linear combination of the first two. That is,

Try showing that a different set, forms a linearly independent set ofsolutions of the differential equation

y999 2 3y0 1 3y9 2 y 5 0.

He x, xe x, x2e xJ,

sx 1 1de x 5 e x 1 xe x.

He x, xe x, sx 1 1de xJ


5 0 W 5 |e xe xe x xe xsx 1 1de xsx 1 2de x sx 1 1de xsx 1 2de xsx 1 3de x|y999 2 3y0 1 3y9 2 y 5 0.

y999 2 3y0 1 3y9 2 y 5 0.


C3C1, C2,

y 5 C1 1 C2 cos x 1 C3 sin x

H1, cos x, sin xJ

W

5 1 5 sin2 x 1 cos2 x

W 5 |100 cos x2sin x2cos x sin xcos x2sin x|y999 1 y9 5 0.

y999 1 y9 5 0.

H1, cos x, sin xJ


CONIC SECTIONS AND ROTATION

Every conic section in the -plane has an equation that can be written in the form

Identifying the graph of this equation is fairly simple as long as the coefficient of the-term, is zero. When is zero, the conic axes are parallel to the coordinate axes, and

the identification is accomplished by writing the equation in standard (completedsquare) form. The standard forms of the equations of the four basic conics are given in the following summary. For circles, ellipses, and hyperbolas, the point is thecenter. For parabolas, the point is the vertex.sh, kd

sh, kd

bxyb,

ax2 1 bxy 1 cy2 1 dx 1 ey 1 f 5 0.xy


Standard Forms of Equations of Conics

Circle

Ellipse

Hyperbola

Parabola

x

Focus( + )h p, k

Vertex( , )h k

p > 0

( ) = 4 ( )y k p x h 2

y

Focus(h, k + p)

Vertex(h, k)

p > 0

(x h)2 = 4p(y k)x

y

sp 5 directed distance from vertex to focus):

(y k)2 (x h)2 = 1

x

y

2

2

(h, k)

2 2

(x h)2 (y k)2 = 1

x

y

2

2

2

(h, k)

2

s2a 5 transverse axis length, 2b 5 conjugate axis lengthd:

(x h)2 (y k)2+ = 1

x

y2

(h, k) 2

2

2

x

(x h)2 (y k)22 2+ = 1

y

(h, k)

2

2

s2a 5 major axis length, 2b 5 minor axis lengthd:sx 2 hd2 1 sy 2 kd2 5 r 2sr 5 radiusd:

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Identifying Conic Sections

a. The standard form of is

The graph of this equation is a parabola with the vertex at The axisof the parabola is vertical. Because the focus is the point Finally,because the focus lies below the vertex, the parabola opens downward, as shown in Figure 4.20(a).

b. The standard form of is

The graph of this equation is an ellipse with its center at The majoraxis is horizontal, and its length is The length of the minor axis is The vertices of this ellipse occur at and and the endpoints of theminor axis occur at and as shown in Figure 4.20(b).

a. b.

Figure 4.20

The equations of the conics in Example 5 have no -term. Consequently, the axesof the graphs of these conics are parallel to the coordinate axes. For second-degreeequations that have an -term, the axes of the graphs of the corresponding conics arenot parallel to the coordinate axes. In such cases it is helpful to rotate the standard axes to form a new -axis and -axis. The required rotation angle (measured counterclockwise) is With this rotation, the standard basis for

is rotated to form the new basis

as shown in Figure 4.21.

Figure 4.21

To find the coordinates of a point relative to this new basis, you can use atransition matrix, as demonstrated in Example 6.

sx, yd

y'

x

x'

(cos , sin )(sin , cos )

(0, 1)

(1, 0)

y

B9 5 Hscos u, sin ud, s2sin u, cos udJ

B 5 Hs1, 0d, s0, 1dJR2,cot 2u 5 sa 2 cdyb.

uy9x9

xy

xy

5 4 3 2 1

1

2

3

x

(3, 2)

(3, 1)(1, 1)(5, 1)

(3, 0)

y

(x + 3)24

(y 1)21+ = 1

2 2 3 4

3

2

1

1

x

(1, 1)(1, 0)Focus

y

(x 1)2 = 4(1)( y 1)

s23, 0d,s23, 2ds21, 1d,s25, 1d

2b 5 2.2a 5 4.sh, kd 5 s23, 1d.

sx 1 3d24

1sy 2 1d2

15 1.

x2 1 4y2 1 6x 2 8y 1 9 5 0

s1, 0d.p 5 21,sh, kd 5 s1, 1d.

sx 2 1d2 5 4s21dsy 2 1d.

x2 2 2x 1 4y 2 3 5 0


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JYWangIf b is nonzero, it is helpful for identifying the graph of a conic section by rotating the coordinate axes such that the conic section is parallel to the new axes

A Transition Matrix for Rotation in

Find the coordinates of a point in relative to the basis

SOLUTION

By Theorem 4.21 you have

Because is the standard basis for is represented by You can use theformula given in Section 2.3 (page 66) for the inverse of a matrix to find This results in

By letting be the coordinates of relative to you can use the transitionmatrix as follows.

The - and -coordinates are and

The last two equations in Example 6 give the -coordinates in terms of the -coordinates. To perform a rotation of axes for a general second-degree equation, it is

helpful to express the -coordinates in terms of the -coordinates. To do this, solvethe last two equations in Example 6 for and to obtain

and

Substituting these expressions for and into the given second-degree equation produces a second-degree equation in and that has no -term.

The proof of the above result is left to you. (See Exercise 82.)

x9y9y9x9yx

y 5 x9 sin u 1 y9 cos u.x 5 x9 cos u 2 y9 sin u

yxx9y9xy

xyx9y9

y9 5 2x sin u 1 y cos u. x9 5 x cos u 1 y sin uy9x9

3 cos u2sin usin ucos u4 3

x

y4 5 3x9

y94P21

B9,sx, ydsx9, y9d

fI P21g 5 3100 1

cos u

2sin u sin ucos u4.

sB9d21.2 3 2sB9d21.P21R2,B

fB9 Bg 5 3cos usin u2sin u

cos u

10

014.

B9 5 Hscos u, sin ud, s2sin u, cos udJ.

R2sx, yd

R2


LINEAR ALGEBRA APPLIED

A satellite dish is an antenna that is designed to transmit

or receive signals of a specific type. A standard satellite

dish consists of a bowl-shaped surface and a feed horn

that is aimed toward the surface. The bowl-shaped surface

is typically in the shape of an elliptic paraboloid. (See

Section 7.4.) The cross section of the surface is typically

in the shape of a rotated parabola.

Rotation of Axes

The general second-degree equation canbe written in the form

by rotating the coordinate axes counterclockwise through the angle where is defined by The coefficients of the new equation are obtainedfrom the substitutions

and y 5 x9 sin u 1 y9 cos u. x 5 x9 cos u 2 y9 sin u

cot 2u 5a 2 c

b.

uu,

a9sx9d2 1 c9sy9d2 1 d9x9 1 e9y9 1 f9 5 0

ax 2 1 bxy 1 cy2 1 dx 1 ey 1 f 5 0

Chris H. Galbraith/Shutterstock.com

JYWangtarget basis

JYWangoriginal basis

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JYWang[B'|B] [I | P^(-1)] vs.[B'| I] [I | B'^(-1)]

JYWangP^(-1) * [v]_B = [v]_B'

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Example 7 demonstrates how to identify the graph of a second-degree equation byrotating the coordinate axes.

Rotation of a Conic Section

Perform a rotation of axes to eliminate the -term in

and sketch the graph of the resulting equation in the -plane.

SOLUTION

The angle of rotation is given by

This implies that So,

and

By substituting

and

into the original equation and simplifying, you obtain

Finally, by completing the square, you find the standard form of this equation to be

which is the equation of an ellipse, as shown in Figure 4.22.

In Example 7 the new (rotated) basis for is

and the coordinates of the vertices of the ellipse relative to are

and

To find the coordinates of the vertices relative to the standard basisuse the equations

and

to obtain and as shown in Figure 4.22.s2!2, 0d,s23!2, 22!2d

y 51!2

sx9 1 y9 d

x 51!2

sx9 2 y9 d

B 5 Hs1, 0d, s0, 1dJ,

32114.325

14B9

B9 5 51 1!2, 1!22, 12

1!2

,

1!226

R2

sx9 1 3d222

1sy9 2 1d2

125

sx9 1 3d24

1sy9 2 1d2

15 1

sx9 d2 1 4sy9 d2 1 6x9 2 8y9 1 9 5 0.

y 5 x9 sin u 1 y9 cos u 51!2

sx9 1 y9d

x 5 x9 cos u 2 y9 sin u 51!2

sx9 2 y9d

cos u 51!2

.sin u 51!2

u 5 py4.

cot 2u 5a 2 c

b5

5 2 526

5 0.

x9y9

5x2 2 6xy 1 5y2 1 14!2x 2 2!2y 1 18 5 0

xy


Figure 4.22

5 4

4

3

2

1

2y'

x'

x

( 2, 0)

(3 2, 2 2)

= 45

y

(x' + 3)24

(y' 1)21+ = 1

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JYWangTwo vertex points on the major axis in the new basis

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JYWangTwo vertex points on the major axis in the original basis

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Applications in Ch4