Section 3.2 WebHW due 9/29 O ce hours Tues, Thurs 1-2 pm

Agenda

Section 3.2

Reminders

Lab 1 write-up due 9/26 or 9/28

Lab 2 prelab due 9/26 or 9/28

WebHW due 9/29

Office hours Tues, Thurs1-2 pm (5852 East Hall)

MathLab office hourSun 7-8 pm (MathLab)

(Gary Marple) September 25th, 2017 Math 216: Introduction to Differential Equations

§3.2 Systems of Two First Order Linear DE’s

Objectives

Be able to write a linear system of DE’s using matrices

Be able to recognize component plots, direction fields,and phase portraits for linear systems

Be able to decide when a linear system is guaranteed tohave unique solutions

Be able to find the critical points of an autonomous linearsystem

Be able to rewrite a 2nd order DE as a 1st order system


Consider two interconnected tanks. Tank 1 initially contains 55 ozof salt, and Tank 2 initially contains 26 oz of salt. Write a systemof DE’s for the amount of salt in each tank.

dQ1

dt=

1.5 gal

min

1 oz

gal+

1.5 gal

min

Q2 oz

20 gal−3 gal

min

Q1 oz

30 gal,

dQ2

dt=

1 gal

min

3 oz

gal+

3 gal

min

Q1 oz

30 gal−4 gal

min

Q2 oz

20 gal,

Q1(0) = 55 oz, Q2(0) = 26 oz.



dQ1

dt=

1.5 gal

min

1 oz

gal+

1.5 gal

min

Q2 oz

20 gal−3 gal

min

Q1 oz

30 gal,

dQ2

dt=

1 gal

min

3 oz

gal+

3 gal

min

Q1 oz

30 gal−4 gal

min

Q2 oz

20 gal,

Q1(0) = 55 oz, Q2(0) = 26 oz.



dQ1

dt=

1.5 gal

min

1 oz

gal+

1.5 gal

min

Q2 oz

20 gal−3 gal

min

Q1 oz

30 gal,

dQ2

dt=

1 gal

min

3 oz

gal+

3 gal

min

Q1 oz

30 gal−4 gal

min

Q2 oz

20 gal,

Q1(0) = 55 oz, Q2(0) = 26 oz.



dQ1

dt=

1.5 gal

min

1 oz

gal+

1.5 gal

min

Q2 oz

20 gal−3 gal

min

Q1 oz

30 gal,

dQ2

dt=

1 gal

min

3 oz

gal+

3 gal

min

Q1 oz

30 gal−4 gal

min

Q2 oz

20 gal,

Q1(0) = 55 oz, Q2(0) = 26 oz.



dQ1

dt=

1.5 gal

min

1 oz

gal+

1.5 gal

min

Q2 oz

20 gal−3 gal

min

Q1 oz

30 gal,

dQ2

dt=

1 gal

min

3 oz

gal+

3 gal

min

Q1 oz

30 gal−4 gal

min

Q2 oz

20 gal,

Q1(0) = 55 oz, Q2(0) = 26 oz.



dQ1

dt=

1.5 gal

min

1 oz

gal+

1.5 gal

min

Q2 oz

20 gal−3 gal

min

Q1 oz

30 gal,

dQ2

dt=

1 gal

min

3 oz

gal+

3 gal

min

Q1 oz

30 gal−4 gal

min

Q2 oz

20 gal,

Q1(0) = 55 oz, Q2(0) = 26 oz.



dQ1

dt=

1.5 gal

min

1 oz

gal+

1.5 gal

min

Q2 oz

20 gal−3 gal

min

Q1 oz

30 gal,

dQ2

dt=

1 gal

min

3 oz

gal+

3 gal

min

Q1 oz

30 gal−4 gal

min

Q2 oz

20 gal,

Q1(0) = 55 oz, Q2(0) = 26 oz.



dQ1

dt=

1.5 gal

min

1 oz

gal+

1.5 gal

min

Q2 oz

20 gal−3 gal

min

Q1 oz

30 gal,

dQ2

dt=

1 gal

min

3 oz

gal+

3 gal

min

Q1 oz

30 gal−4 gal

min

Q2 oz

20 gal,

Q1(0) = 55 oz, Q2(0) = 26 oz.



dQ1

dt=

1.5 gal

min

1 oz

gal+

1.5 gal

min

Q2 oz

20 gal−3 gal

min

Q1 oz

30 gal,

dQ2

dt=

1 gal

min

3 oz

gal+

3 gal

min

Q1 oz

30 gal−4 gal

min

Q2 oz

20 gal,

Q1(0) = 55 oz, Q2(0) = 26 oz.


Simplifying our expression gives

dQ1

dt= −0.1Q1 + 0.075Q2 + 1.5,

dQ2

dt= 0.1Q1 − 0.2Q2 + 3,

Q1(0) = 55, Q2(0) = 26.

Let

q =

[Q1

Q2

], b =

[1.53

], A =

[−0.1 0.0750.1 −0.2

].

We can express the system, using matrix notation, as

dq

dt= Aq + b, q(0) =

[5526

].


The solution of the system turns out to be

q =

[Q1(t)Q2(t)

]= 7

[1−2

]e−t/4 + 2

[32

]e−t/20 +

[4236

].

Combining terms gives us equations for Q1(t) and Q2(t).

Q1(t) = 7e−t/4 + 6e−t/20 + 42,

Q2(t) = −14e−t/4 + 4e−t/20 + 36.

We can visualize Q1 and Q2 by plotting both functions on thesame graph. Plots of Q1 and Q2 versus t are calledcomponent plots.

(Note that this solution was given. We have not yet discussedhow to solve such a system.)


Component Plot for Q1(t) and Q2(t)

Q1,Q2 ↑ t →


In our example, the right-hand side of the DE only involved thedependent variables Q1 and Q2. Such a system is calledautonomous, just like it was when we had individual DE’s.Another way to visualize solutions of 2× 2 autonomous systems isto consider plots with Q1 on the x-axis and Q2 on the y -axis. Thetwo dependent variables Q1 and Q2 are sometimes referred to asstate variables since the state of the system at any time dependson their values. In addition, the Q1Q2-plane is sometimes referredto as state space, the state plane, or the phase plane. Thereare two common plots in state space. The first one, called avector field, is similar to a slope field, except each line segment isa normalized vector (an arrow) pointing in the direction of thevector 〈dQ1/dt, dQ2/dt〉, the slope of which can be easilycomputed using the chain rule. That is,

dQ2

dQ1

dQ1

dt=

dQ2

dt.

The second plot, called a phase portrait, is the 2-dimensionalversion of a phase line.


Vector Field showing Q2 vs Q1

Q2 ↑ Q1 →The solid line corresponds to the solution with initial condition

Q1(0) = 55, Q2(0) = 26.(Gary Marple) September 25th, 2017 Math 216: Introduction to Differential Equations

Phase Portrait showing Q2 vs Q1


Example

Find the equilibrium solutions of the linear system

q′ = Aq + b,

where

q =

[Q1

Q2

], A =

[−0.1 0.0750.1 −0.2

], b =

[1.53

]Remember that equilibrium solutions are constant solutions.Therefore, we need to set q′ = 0.

0 = Aqeq + b

Aqeq = −b (1)

We now have a linear system of equations. We can write thesystem as an augmented matrix. That is,


Example

Find the equilibrium solutions of the linear system

q′ = Aq + b,

where

q =

[Q1

Q2

], A =

[−0.1 0.0750.1 −0.2

], b =

[1.53

]Remember that equilibrium solutions are constant solutions.Therefore, we need to set q′ = 0.

0 = Aqeq + b

Aqeq = −b (2)

We now have a linear system of equations. We can write thesystem as an augmented matrix. That is,


[−0.1 0.075 −1.50.1 −0.2 −3

]Add R1 to R2[

−0.1 0.075 −1.50 −0.125 −4.5

]The matrix is now in row echelon form. We can see thatcolumns 1 and 2 are pivot columns, so Q1 and Q2 are basicvariables. Therefore, we will try to solve for both variables.Converting back to a linear system gives us

−0.1Q1 + 0.075Q2 = −1.5,

0Q1 − 0.125Q2 = −4.5.

We can use the last equation to solve for Q2. That is,

Q2 = 4.5/0.125 = 36.


[−0.1 0.075 −1.50.1 −0.2 −3

]Add R1 to R2[

−0.1 0.075 −1.50 −0.125 −4.5


−0.1Q1 + 0.075Q2 = −1.5,

0Q1 − 0.125Q2 = −4.5.


Q2 = 4.5/0.125 = 36.


[−0.1 0.075 −1.50.1 −0.2 −3

]Add R1 to R2[

−0.1 0.075 −1.50 −0.125 −4.5


−0.1Q1 + 0.075Q2 = −1.5,

0Q1 − 0.125Q2 = −4.5.


Q2 = 4.5/0.125 = 36.


[−0.1 0.075 −1.50.1 −0.2 −3

]Add R1 to R2[

−0.1 0.075 −1.50 −0.125 −4.5


−0.1Q1 + 0.075Q2 = −1.5,

0Q1 − 0.125Q2 = −4.5.


Q2 = 4.5/0.125 = 36.


[−0.1 0.075 −1.50.1 −0.2 −3

]Add R1 to R2[

−0.1 0.075 −1.50 −0.125 −4.5


−0.1Q1 + 0.075Q2 = −1.5,

0Q1 − 0.125Q2 = −4.5.


Q2 = 4.5/0.125 = 36.


[−0.1 0.075 −1.50.1 −0.2 −3

]Add R1 to R2[

−0.1 0.075 −1.50 −0.125 −4.5


−0.1Q1 + 0.075Q2 = −1.5,

0Q1 − 0.125Q2 = −4.5.


Q2 = 4.5/0.125 = 36.


−0.1Q1 + 0.075Q2 = −1.5,

0Q1 − 0.125Q2 = −4.5.

Plugging Q2 = 36 into the first equation and solving for Q1

givesQ1 = (1.5/0.1) + (0.075 · 36)/0.1 = 42.

Therefore,

qeq =

[4236

].


−0.1Q1 + 0.075Q2 = −1.5,

0Q1 − 0.125Q2 = −4.5.


givesQ1 = (1.5/0.1) + (0.075 · 36)/0.1 = 42.

Therefore,

qeq =

[4236

].


−0.1Q1 + 0.075Q2 = −1.5,

0Q1 − 0.125Q2 = −4.5.


givesQ1 = (1.5/0.1) + (0.075 · 36)/0.1 = 42.

Therefore,

qeq =

[4236

].


Example

Find the critical points of

x′ = Ax + b,

where

x =

[x1

x2

], A =

[2 34 6

], b =

[12

].

Recall that a critical point is the same thing as an equilibriumsolution. Therefore, we’ll set x′ = 0.

0 = Axeq + b

Axeq = −b

We now have a linear system we can express as an augmentedmatrix.


Example

Find the critical points of

x′ = Ax + b,

where

x =

[x1

x2

], A =

[2 34 6

], b =

[12

].

Recall that a critical point is the same thing as an equilibriumsolution. Therefore, we’ll set x′ = 0.

0 = Axeq + b

Axeq = −b

We now have a linear system we can express as an augmentedmatrix.


[2 3 −14 6 −2

]Add − 2R1 to R2[

2 3 −10 0 0

]The matrix is now in row echelon form. In this case, only thefirst column is a pivot column. Therefore, x1 is a basic variableand x2 is a free variable. We’ll start by setting x2 equal to aparameter. That is,

x2 = c .

We can write the row reduced matrix as a linear system. Thatis,

2x1 + 3x2 = −1,

0x1 + 0x2 = 0.


[2 3 −14 6 −2

]Add − 2R1 to R2[

2 3 −10 0 0


x2 = c .


2x1 + 3x2 = −1,

0x1 + 0x2 = 0.


[2 3 −14 6 −2

]Add − 2R1 to R2[

2 3 −10 0 0


x2 = c .


2x1 + 3x2 = −1,

0x1 + 0x2 = 0.


[2 3 −14 6 −2

]Add − 2R1 to R2[

2 3 −10 0 0


x2 = c .


2x1 + 3x2 = −1,

0x1 + 0x2 = 0.


[2 3 −14 6 −2

]Add − 2R1 to R2[

2 3 −10 0 0


x2 = c .


2x1 + 3x2 = −1,

0x1 + 0x2 = 0.


[2 3 −14 6 −2

]Add − 2R1 to R2[

2 3 −10 0 0


x2 = c .


2x1 + 3x2 = −1,

0x1 + 0x2 = 0.


2x1 + 3x2 = −1,

0x1 + 0x2 = 0

If we look at the first equation, plug c in for x2, and solve forx1, we get

x1 = −1

2− 3

2c .

Therefore,

xeq =

[x1

x2

]=

[−1

2− 3

2c

c

]=

[−1/2

0

]+ c

[−3/2

1

].

In other words, every point on a line in the x1x2-plane is acritical point. Such a situation arises because the two DE’s areexactly the same, just off by a constant.


2x1 + 3x2 = −1,

0x1 + 0x2 = 0


x1 = −1

2− 3

2c .

Therefore,

xeq =

[x1

x2

]=

[−1

2− 3

2c

c

]=

[−1/2

0

]+ c

[−3/2

1

].



2x1 + 3x2 = −1,

0x1 + 0x2 = 0


x1 = −1

2− 3

2c .

Therefore,

xeq =

[x1

x2

]=

[−1

2− 3

2c

c

]=

[−1/2

0

]+ c

[−3/2

1

].



Example

Rewrite the second order DE as a first order system.

2d2y

dt2− 0.5

dy

dt+ 8y = 6 sin (2t)

If we let x = dy/dt, then dx/dt = d2y/dt2 and

dx

dt=

1

4x − 4y + 3 sin (2t)

dy

dt= x

We can express this system of DE’s in the formx′ = Ax + b(t). That is,

d

dt

[xy

]=

[1/4 −4

1 0

] [xy

]+

[3 sin (2t)

0

].


Example


2d2y

dt2− 0.5

dy

dt+ 8y = 6 sin (2t)


dx

dt=

1

4x − 4y + 3 sin (2t)

dy

dt= x


d

dt

[xy

]=

[1/4 −4

1 0

] [xy

]+

[3 sin (2t)

0

].


Example


2d2y

dt2− 0.5

dy

dt+ 8y = 6 sin (2t)


dx

dt=

1

4x − 4y + 3 sin (2t)

dy

dt= x


d

dt

[xy

]=

[1/4 −4

1 0

] [xy

]+

[3 sin (2t)

0

].


Example


t2u′′ + tu′ + (t2 − 0.25)u = 0

If we let v = du/dt. Then, dv/dt = d2u/dt2 and

u′ = v

v ′ = −1

tv − t2 − 0.25

t2u

We can express the system in the form u′ = A(t)u. That is,

d

dt

[uv

]=

[0 1

(0.25− t2)/t2 −1/t

] [uv

].


Example


t2u′′ + tu′ + (t2 − 0.25)u = 0


u′ = v

v ′ = −1

tv − t2 − 0.25

t2u


d

dt

[uv

]=

[0 1

(0.25− t2)/t2 −1/t

] [uv

].


Example


t2u′′ + tu′ + (t2 − 0.25)u = 0


u′ = v

v ′ = −1

tv − t2 − 0.25

t2u


d

dt

[uv

]=

[0 1

(0.25− t2)/t2 −1/t

] [uv

].


Theorem: Existence and Uniqueness of Solutions

Let each of the functions p11, . . . , p22, g1, and g2 becontinuous on an open interval I = α < t < β. Let t0 be anypoint in I , and let x0 and y0 be any given numbers. Then,there exists a unique solution to the IVP

dx

dt= P(t)x + g(t), x(t0) =

[x0

y0

],

where

x =

[xy

], P(t) =

[p11(t) p12(t)p21(t) p22(t)

], g(t) =

[g1(t)g2(t)

].

Furthermore, the solution exists throughout the interval I .


Example

Transform the given IVP into an IVP with two first orderequations. Then write the system in matrix form.

tu′′ + u′ + tu = 0, u(1) = 1, u′(1) = 0



Example

Find the critical points of the system of DE’s.

x ′ = −x + y + 1, y ′ = x + y − 3



Documents

Section 3.2 WebHW due 9/29 O ce hours Tues, Thurs 1-2 pm