Math 331 Homework Assignment Chapter 7 Page 1 of 9 ...

Math 331

Homework Assignment Chapter 7

Page 1 of 9

Instructions: Please make sure to demonstrate every step in your calculations. Return your answers including this homework sheet back to the instructor as a single, stapled package. Also, please keep a copy of your solutions for your reference, especially in view of studying for exams.

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1. Write the following differential equations as first order systems. State whether the system is linear or nonlinear. (a) y + 4t2y - y = 0 (b) y + y - y2 = 0 (c) y - 2y = 5e-t

Solution: (a) First rewrite the differential equation as y = y - 4t2y . Now define x1 =

y, x2 = y and x3 = y . Then we have the system

x1 = (y) = x2 x2 = (y ) = x3 x3 = (y ) = y - 4t2y = x2 - 4t2x3

This is a linear system with the coefficient matrix

0 1 0 P(t) = 0 0 1

0 1 -4t2

and g(t) = 0.

(b) Define x1 = y and x2 = y . Then we have

This is a nonlinear system.

x1 = x2 x2 = x21 - x2

(c) Define x1 = y, x2 = y , x3 = y and x4 = y . Then we have the system

x1 = x2 x2 = x3 x3 == x4 x4 = 2x1 + 5e-t

Math 331

Homework Assignment Chapter 7

Page 2 of 9

This is a linear system with

0 1 0 0

P(t)

=

0 0

0 0

1 0

0 1

,

2000

0

g(t)

=

0 0

5e-t

2. Let x = x1(t), y = y1(t) and x = x2(t), y = y2(t) be any two solutions of the linear nonhomogeneous system

x = p11(t) x + p12(t) y + g1(t), y = p21(t) x + p22(t) y + g2(t).

Show that x = x1(t) - x2(t), y = y1(t) - y2(t) is a solution of the corresponding homogeneous system.

Solution: To show this, let us consider the homogeneous system of Eq. (7), i.e., with g1(t) = g2(t) = 0 t. Then, direct substitution of x = x1(t) - x2(t), y = y1(t) - y2(t) into the homogeneous system yields to

(x1 - x2) = p11(t) (x1 - x2) + p12(t) (y1 - y2), (y1 - y2) = p21(t) (x1 - x2) + p22(t) (y1 - y2),

and upon reshuffling terms we obtain

x1 - (p11(t)x1 + p12(t)y1) = x2 - (p11(t)x2 + p12(t)y2) , y1 - (p21(t)x1 + p22(t)y1) = y2 - (p21(t)x2 + p22(t)y2) .

Next, note that the left- and right-hand-sides equal to g1(t) and g2(t) for the first and second equations, respectively (recall that x = x1(t), y = y1(t) and x = x2(t), y = y2(t) are solutions to the nonhomogeneous system). This way, we conclude that x = x1(t) - x2(t), y = y1(t) - y2(t) is indeed a solution of the corresponding homogeneous system.

3. For the following matrices compute the eigenvalues and eigenvectors and indicate the type of the system (saddle point, node (or sink), source, center, spiral source, spiral sink).

(a)

A=

2 -1

-1 2

(b)

A=

0 2

-2 0

Math 331

Homework Assignment Chapter 7

Page 3 of 9

(c)

A=

3 1

-4 3

Solution: Recall that we determine the eigenvalues via the characteristic equa-

tion:

det (A - rI) = 0.

(1)

Then, upon finding the eigenvalues, we will determine the components of the

eigenvectors, i.e., x1 and x2 of x =

x1 x2

(a) The eigenvalues of the first matrix are r1 = 1 and r2 = 3. For r1 the eigenvector can be found as follows:

2 -1 -1 2

x1 x2

=

x1 x2

x1 = x2.

Thus, for x1 = 1, the associated eigenvector is x(1) =

1 1

.

On equally

footing and for r2 at hand, we have

2 -1 -1 2

x1 x2

=3

x1 x2

-x1 = x2.

For x1 = 1, we obtain x(2) =

1 -1

.

(b) The eigenvalues of the second matrix are r1 = 2i and r2 = -2i. The associated eigenvector x(1) for r1 can be obtained via

0 -2 20

x1 x2

= 2i

x1 x2

-2x2 = 2ix1.

For x1 = 1 we obtain x(1) =

1 -i

. Finally, and as per the second eigenvalue

2, we have similarly

0 -1 10

x1 x2

= -i

x1 x2

x1 = -ix2,

and for x1 = 1, the eigenvector is x(2) =

1 i

.

(c) The eigenvalues of the third matrix are r1 = 3 + 2i and r2 = 3 - 2i. The associated eigenvector x(1) for r1 can be obtained via

3 -4 13

x1 x2

= (3 + 2i)

x1 x2

2ix2 = x1.

For x2 = 1 we obtain x(1) =

(2i 1 . Finally for the second eigenvalue 2,

the eigenvector is x(2) =

-2i i

.

Math 331

Homework Assignment Chapter 7

Page 4 of 9

4. For the system of differential equations

x = Ax

A=

2 -5 1 -2

and the two vector-valued functions

x1(t) =

5 cos t 2 cos t + sin t

,

x2(t) =

5 sin t 2 sin t - cos t

do the following:

(a) Show that the given functions are solutions of the given system of differential equations.

(b) Show that x = c1x1 + c2c2 is also a solution of the given system for any value of c1 and c2.

(c) Find the solution of the given system that satisfies the initial condition x(0) = (1, 2)T

Solution: (a) Compute

x1 =

-5 sin t -2 sin t - cos t

,

x2 =

5 cos t 2 cos t + sin t

and

Ax1 =

10 cos t - 10 cos t - 5 sin t 5 cos t - 4 cos t - 2 sin t

,

Ax2 =

10 sin t - 10 sin t + 5 cos t 5 sin t - 4 sin t + 2 cos t

Upon simplifying, we clearly have the equalities Ax1 = x1 and Ax2 = x2. (b) Using (a), we have

Ax = A(c1x1 + c2x2) = c1Ax1 + c2Ax2 = c1x1 + c2x2 = (c1x1 + c2x2) =x

(c) We need to find c1 and c2 such that

c1x1(0) + c2x2(0) =

1 2

This gives us the two equations

5c1 = 1, 2c1 - c2 = 2

which has solution c1 = 1/5 and c2 = -8/5. So 18

x(t) = 5 x1 - 5 x2

Math 331

Homework Assignment Chapter 7

Page 5 of 9

5. Solve the given initial value problem. Describe the behavior of the solution as t

x=

5 3

-1 1

x,

x(0) =

2 -1

Solution: Firstly, finding the eigenvalues and eigenvectors of the coefficient ma-

trix A as follows:

det (A - rI) =

5-r 3

-1 1-r

= 0 r2 - 6r + 8 = (r - 2)(r - 4) = 0

The two eigenvalues are r1 = 2 and r2 = 4.

For r1 = 2, the corresponding eigenvector (1) = (1(1), 2(1))T can be found by solving

5 - 2 -1 3 1-2

1(1) 2(1)

=

0 0

Hence, 31(1) = 2(1). So the eigenvector (1) corresponding to the eigenvalue r1 = 2

is

(1) =

1 3

Similarly, for r2 = 4, the corresponding eigenvector (2) = (1(2), 2(2))T can be found by solving

5 - 4 -1 3 1-4

1(2) 2(2)

=

0 0

Hence, 1(2) = 2(2). So the eigenvector (2) corresponding to the eigenvalue r2 = 4

is

(2) =

1 1

Thus a fundamental set of solutions of the system is

x(1)(t) =

1 3

e2t,

x(2)(t) =

1 1

e4t

The general solution is

x(t) = c1x(1)(t) + c2x(2)(t) = c1

1 3

e2t + c2

1 1

e4t

Plug in initial value x(0) = (2, -1)T and solve for c1 and c2

3

7

c1

=

-, 2

c2 = 2

Thus the particular solution for the initial value problem is

3 x(t) = -

2

1 3

e2t + 7 2

1 1

e4t

As

t

,

the

second

term

7 2

1 1

e4t

is

dominant

and

the

first

term

-

3 2

1 3

e2t

is negligible. x(t) becomes asymptotic to the line x2 = x1 as t

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