Section 5.1 First-Order Systems & Applications

Section 5.1First-Order Systems & Applications

Suppose x and y are both functions of t. Solve: x′ = 3x – yy′ = 2x + y – et

Why would we consider such things?

Ex. 1 Give the system of diff eqs which describe the following tanks containing brine solution:

Theorem:Consider the following system of diff eqs (all xi, pij, and fi are functions of t )

x1′ = p11x1 + p12x2 + p13x3 + + ⋯ p1nxn + f1 x2′ = p21x1 + p22x2 + p23x3 + + ⋯ p2nxn + f2 x3′ = p31x1 + p32x2 + p33x3 + + ⋯ p3nxn + f3 : :xn′ = pn1x1 + pn2x2 + pn3x3 + + ⋯ pnnxn + fn

Let J be an open interval containing t = a. Suppose the functions pij and the functions fk are continuous on J. Then the system of differential equations has a unique solution that satisfies the following initial conditions: x1(a) = b1 , x2(a) = b2 , x3(a) = b3 , ......... , xn(a) = bn

If we have a system of higher order diff eqs, we can transform it into a system of first order diff eqs.

Ex. 3 Rewrite the one diff eq x(3) + 3x″ + 2x′ – 5x = sin(2t) as a system of first order diff eqs.

Ex. 4 Rewrite the following system as a system of first order diff eqs.2x″ = –6x + 2yy″ = 2x – 2y + 40sin(3t)

Section 5.2The Method Of Elimination

Review of solving systems of algebraic equations (2 equations with 2 unknowns) i. Substitution method ii. Elimination method iii. Cramer's rule

Ex. 1 Find the general solution to the following system by using a variant of the substitution method.

x′ = 4x – 3yy′ = 6x – 7y

Ex. 2 Find the general solution to the following system by using a variant of the elimination method.

x′ = 4x – 3yy′ = 6x – 7y

We shall now use the following notation:

L will be a linear operator of the form L = anDn + an–1Dn–1 + an–2Dn–2 + + ⋯ a2D2 + a1D + a0

Ex. 3 Find the general solution to the following general system of diff eqs: L1x + L2y = f1(t)L3x + L4y = f2(t)

Ex. 4 Find the general solution to x′ = 2x + yy′ = 2x + 3y + e5t

Solution: (D–2)x – y = 0 –2x + (D–3)y = e5t

[(D–2)(D–3) – 2] x = (D – 3)0 + e5t [(D–2)(D–3) – 2] y = (D – 2) e5t + 2(0)

(D2–5D+4) x = e5t (D2–5D+4) y = 5e5t – 2e5t

(D–4)(D–1) x = e5t (D–4)(D–1) y = 3e5t

xc = c1e4t + c2et yc = c3e4t + c4et

xp = Ae5t ⇒ xp = (1∕4)e5t yp = Be5t ⇒ yp = (3∕4)e5t

x = c1e4t + c2et + (1∕4)e5t y = c3e4t + c4et + (3∕4)e5t

5

2 1 0 12 3 3t

Dx

D e D

5

2 1 2 02 3 2 t

D Dy

D e

Ex. 4 Find the general solution to x′ = 2x + yy′ = 2x + 3y + e5t

Solution:

x = c1e4t + c2et + (1∕4)e5t y = c3e4t + c4et + (3∕4)e5t

Plugging these solutions into the first equation in the initial problem we get:

x′ = 2x + y

(c1e4t + c2et + (1∕4)e5t)′ = 2(c1e4t + c2et + (1∕4)e5t) + (c3e4t + c4et + (3∕4)e5t)

4c1e4t + c2et + (5/4)e5t = 2c1e4t + 2c2et + (1/2)e5t + c3e4t + c4et + (3/4)e5t

0 = (–2c1 + c3)e4t + (c2 + c4)et

–2c1 + c3 = 0 ⇒ c3 = 2c1

c2 + c4 = 0 c4 = –c2

Section 5.3Matrices & Linear Systems

x1′ = 2x1 + x2

x2′ = 2x1 + 3x2

x1′ = 2x1 + x2 ⇒x2′ = 2x1 + 3x2

2 12 3

x x

Previously we would state:

The general solution to the system x1′ = 2x1 + x2 turns out to be: x1 = c1e4t + c2et x2′ = 2x1 + 3x2 x2 = 2c1e4t – c2et



We now would state:

The general solution to the system turns out to be:

2 12 3

x x

4

1 242

t t

t t

e ex c c

e e



We now would state:

The general solution to the system turns out to be:

We shall find that general solutions to these "2x2 homogenous systems" will take this form of (where and are two linearly independent vectors).

2 12 3

x x

4

1 242

t t

t t

e ex c c

e e

(1) (2)1 2x c x c x

(1)x (2)x


x Px f


This system of diff eqs is said to be homogenous if

x Px f

0f


This system of diff eqs is said to be homogenous if

Thus, the matrix equation would be for a homogenous system.

x Px f

0f

x Px

We shall now see many theorems very similar to theorems and definitions we had in chapter 2.

Definition: are said to be linearly independent if the equation

only has the solution of c1 = c2 = = ⋯ cn = 0.

(1) (2) (3) ( ), , , .... , nx x x x

(1) (3) ( )1 2 3 0n

nc x c x c x c x

Definition:If are solutions to then we define the Wronskian of these solution to be

(1) (2) (3) ( ), , , .... , nx x x x x Px

(1) (2) (3) ( ) (1) (2) (3) ( )

(1) (2) (3) ( )1 1 1 1(1) (2) (3) ( )2 2 2 2(1) (2) (3) ( )3 3 3 3

(1) (2) (3) ( )

, , , .... , , , , .... ,n n

n

n

n

nn n n n

W x x x x x x x x

x x x xx x x xx x x x

x x x x

Theorem:Suppose are solutions to on an interval J where all the pij functions are continuous.

(a) If are linearly dependent then W = 0 for every point on the interval J.

(b) If are linearly independent then W ≠ 0 for every point on the interval J.

(1) (2) (3) ( ), , , .... , nx x x x x Px

(1) (2) (3) ( ), , , .... , nx x x x

(1) (2) (3) ( ), , , .... , nx x x x

Theorem:Suppose are linearly independent solutions to on an interval J where all the pij functions are continuous. The general solution to is given as:

(1) (2) (3) ( ), , , .... , nx x x x x Px

x Px

(1) (3) ( )1 2 3

nnx c x c x c x c x

Theorem:Suppose is a particular solution to the nonhomogenous system and is the general solution to the corresponding homogenous system . Then the general solution to the nonhomogenous system is .

px x Px f

cx x Px

x Px f

p cx x x

Ex. 1 x′1 = x1 + x2 – 2x3 x′2 = –x1 + 2x2 + x3 x′3 = x2 – x3

(a) Write this system as a matrix equation.

Ex. 1 x′1 = x1 + x2 – 2x3 x′2 = –x1 + 2x2 + x3 x′3 = x2 – x3

(b) Verify that the following three vectors are solution vectors:

2

(1) (2) (3) 2

2

30 , 2 , 3

t t t

t t

t t t

e e ex x e x e

e e e

Ex. 1 x′1 = x1 + x2 – 2x3 x′2 = –x1 + 2x2 + x3 x′3 = x2 – x3

(c) Verify that these are linearly independent vectors.

2

(1) (2) (3) 2

2

30 , 2 , 3

t t t

t t

t t t

e e ex x e x e

e e e

Ex. 1 x′1 = x1 + x2 – 2x3 x′2 = –x1 + 2x2 + x3 x′3 = x2 – x3

(d) Give the general solution for the system.

Ex. 1 x′1 = x1 + x2 – 2x3 x′2 = –x1 + 2x2 + x3 x′3 = x2 – x3

(e) Give the general solution for x1.

Give the general solution for x2.

Give the general solution for x3.

Ex. 1 x′1 = x1 + x2 – 2x3 x′2 = –x1 + 2x2 + x3 x′3 = x2 – x3

(f) Solve the initial value problem: x′1 = x1 + x2 – 2x3 x1(0) = 9, x2(0) = –2, x3(0) = 5 x′2 = –x1 + 2x2 + x3 x′3 = x2 – x3

Section 5.4The Eigenvalue Method for

Homogeneous Systems

Review of eigenvalues and eigenvectors:

Let A be a square matrix. The vector is said to be an eigenvector for the eigenvalue λ if .

x

Ax x

Review of eigenvalues and eigenvectors:

Let A be a square matrix. The vector is said to be an eigenvector for the eigenvalue λ if .

What is the connection between eigenvectors and solutions to systems of diff eqs?

x

Ax x

Ex. 1 We previously (in section 5.3) found three linearly independent solution

vectors to the system One of these was

Verify that this solution is an eigenvector of

1 1 21 2 1 .

0 1 1x x

2

2

2

3 .

t

t

t

eee

1 1 21 2 1 .

0 1 1

Ex. 2 Suppose that (where A, B, C, and k are constants) is a solution

vector to the system . Show that this solution vector is an eigenvector of the matrix P.

kt

kt

kt

AeBeCe

x Px

Theorem:Given a homogenous system , suppose λ is an eigenvalue of P with eigenvector . Then is a nontrivial solution vector to .

x Px

v tvex Px

Ex. 3 Use eigenvalues/eigenvectors to find the general solution to the following system and write your final solution in scalar form. x′1 = 3x1 + x2 x′2 = 3x1 + 5x2

Note:Suppose where P is an nxn matrix. If there are n distinct real eigenvalues of P then we've got n linearly independent solution vectors. This means we can write down the general solution as

(here the λi are the eigenvalues, the are the eigenvectors and the ci are arbitrary constants).

If we have less than n distinct eigenvalues, or if some of the eigenvalues are complex then we will run into trouble and need to do something else.

x Px

31 21 1 2 2 3 3

nt tt tn nx c v e c v e c v e c v e

iv

Theorem:Given a homogenous system , suppose α + βi is a complex eigenvalue of P with eigenvector . Then the real and imaginary parts of the vector will form two linearly independent solution vectors.

x Px

a bi a bi

ie a bi

Theorem:Given a homogenous system , suppose α + βi is a complex eigenvalue of P with eigenvector . Then the real and imaginary parts of the vector will form two linearly independent solution vectors. Note that there are formulas for these two vectors, but you should not use them!

Instead of using these formulas we shall just use this theorem which indicates that we should find the one vector , then split it up into its real and imaginary parts.

x Px

a bi a bi

ie a bi

1 cos sintv e a t b t

2 sin costv e a t b t

ie a bi

Ex. 4 Use eigenvalues/eigenvectors to find the general solution to x′1 = –x1 – x2 x′2 = 4x1 – x2

Section 5.5Multiple Eigenvalue Solutions

If we have less than n distinct eigenvalues (where P is an nxn matrix) then we may have a hard time finding enough linearly independent solution vectors. When an eigenvalue, say λ, is a repeated root with multiplicity k of the characteristic equation, then two possibilities arise:

If we have less than n distinct eigenvalues (where P is an nxn matrix) then we may have a hard time finding enough linearly independent solution vectors. When an eigenvalue, say λ, is a repeated root with multiplicity k of the characteristic equation, then two possibilities arise:

1. There are "just enough" linearly independent eigenvectors associated with . Then there are "just enough" solution vectors: for us to form the general solution to the diff eqs.

2. There are not enough linearly independent eigenvectors associated with λ. In this case λ is said to be defective. We shall cover the first of these two cases.

1 2 3: , , , , kv v v v

1 2 3, , , ,t t t tkv e v e v e v e

Ex. 1 Use eigenvalues/eigenvectors to find the general solution to x′1 = –13x1 + 40x2 – 48x3 x′2 = –8x1 + 23x2 – 24x3 x′3 = 3x3

Section 5.1 First-Order Systems & Applications

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