ODE-Project Complex Eigenvalues

Section 5.4 Complex Eigenvalues

Objectives

To understand and be able to apply Euler’s formula,

$e^{i β t} = \cos β t + i \sin β t .$
To understand that if a $2 \times 2$ matrix $A$ has two complex eigenvalues, $λ = α \pm i β,$ then the general solution will involve sines and cosines. Furthermore, the origin will be a spiral sink, a spiral source or a center.

Consider the following system,

\begin{matrix} (5.4.1) & (\begin{matrix} d x / d t \\ d y / d t \end{matrix}) = (\begin{matrix} - 3 & 1 \\ - 2 & - 1 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) \end{matrix}

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The characteristic polynomial of the system (5.4.1) is

λ^{2} + 4 λ + 5 .

The roots of this polynomial are

λ_{1} = - 2 + i

and

λ_{2} = - 2 - i

with eigenvectors

v_{1} = (1, 1 + i)

and

v_{2} = (1, 1 - i),

respectively. It is clear from the phase portrait of the system that there are no straight-line solutions (Figure 5.4.1). However, we would like to have real solutions for a linear system with real coefficients.

described in detail following the image — 🔗
Figure 5.4.1. A system with no straight-line solutions

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Subsection 5.4.1 Complex Eigenvalues

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Suppose that we have the system

(\begin{matrix} d x / d t \\ d y / d t \end{matrix}) = (\begin{matrix} 0 & β \\ - β & 0 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = A (\begin{matrix} x \\ y \end{matrix}),

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where

β \neq 0 .

The characteristic polynomial of this system is

det (A - λ I) = λ^{2} + β^{2},

and so we have imaginary eigenvalues

\pm i β .

To find the eigenvector corresponding to

λ = i β,

we must solve the system

(\begin{matrix} - i β & β \\ - β & - i β \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix});

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however, this reduces to solving the equation

i β x = β y .

Thus, we can find a complex eigenvector

(1, i) .

Consequently,

x (t) = e^{i β t} (\begin{matrix} 1 \\ i \end{matrix})

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is a complex solution to the system

x^{'} = A x .

The problem is that we have a real system of differential equations and would like real solutions. We can remedy the situation if we use Euler’s formula,

If you are unfamiliar with Euler’s formula, try expanding both sides as a power series to check that we do indeed have a correct identity.

e^{i β t} = \cos β t + i \sin β t .

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Let us rewrite our solution as

\begin{aligned} x (t) & = e^{i β t} (\begin{array}{c} 1 \\ i \end{array}) \\ = (\begin{array}{c} \cos β t + i \sin β t \\ i (\cos β t + i \sin β t) \end{array}) \\ = (\begin{array}{c} \cos β t + i \sin β t \\ - \sin β t + i \cos β t \end{array}) \\ = (\begin{array}{c} \cos β t \\ - \sin β t \end{array}) + i (\begin{array}{c} \sin β t \\ \cos β t \end{array}) \end{aligned}

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and consider the real and imaginary parts of the solution:

x_{Re} = (\begin{matrix} \cos β t \\ - \sin β t \end{matrix}) and x_{Im} = (\begin{matrix} \sin β t \\ \cos β t \end{matrix}) .

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Since

\begin{aligned} x_{Re}^{'} (t) + i x_{Im}^{'} (t) & = x^{'} (t) \\ = A x (t) \\ = A (x_{Re} (t) + i x_{Im} (t)) \\ = A x_{Re} (t) + i A x_{Im} (t) . \end{aligned}

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we know that

x_{Re}^{'} (t) = A x_{Re} (t)

and

x_{Im}^{'} (t) = A x_{Im} (t)

by setting the real and imaginary parts equal. Thus, both

x_{Re} (t)

and

x_{Im} (t)

are solutions to our system. Moreover, since

x_{Re} (0) = (\begin{matrix} 1 \\ 0 \end{matrix}) and x_{Im} (0) = (\begin{matrix} 0 \\ 1 \end{matrix}),

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we know that the appropriate linear combination of

x_{Re} (t)

and

x_{Im} (t)

will provide a solution to any initial value problem.

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We claim that

\begin{matrix} (5.4.2) & x (t) = c_{1} x_{Re} (t) + c_{2} x_{Im} (t) \end{matrix}

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is a general solution to our system. That is, we must be able to write every solution of our system as a linear combination of

x_{Re} (t)

and

x_{Im} (t) .

y (t) = (\begin{matrix} u (t) \\ v (t) \end{matrix})

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is another solution to the system

x^{'} = A x,

then

y^{'} (t) = (\begin{matrix} u^{'} (t) \\ v^{'} (t) \end{matrix}) = (\begin{matrix} 0 & β \\ - β & 0 \end{matrix}) (\begin{matrix} u (t) \\ v (t) \end{matrix}) = (\begin{matrix} β v (t) \\ - β u (t) \end{matrix}) .

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In other words,

u^{'} (t) = β v (t)

and

v^{'} (t) = - β u (t) .

Now, define

f

f (t) = (u (t) + i v (t)) e^{i β t} .

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The derivative of

f

\begin{aligned} f^{'} (t) & = (u^{'} (t) + i v^{'} (t)) e^{i β t} + i β (u (t) + i v (t)) e^{i β t} \\ = (β v (t) - i β u (t)) e^{i β t} + (i β u (t) + i^{2} β v (t)) e^{i β t} \\ = 0. \end{aligned}

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Therefore,

f (t)

is a complex constant and

f (t) = (u (t) + i v (t)) e^{i β t} = a + b i .

We can now write

u (t) + i v (t) = (a + i b) e^{- i β t} .

Thus,

\begin{aligned} u (t) + i v (t) & = (a + i b) e^{- i β t} \\ = (a + i b) (\cos β t - i \sin β t) \\ = (a \cos β t + b \sin β t) + i (b \cos β t - a \sin β t) . \end{aligned}

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Therefore,

\begin{aligned} u (t) & = a \cos β t + b \sin β t \\ v (t) & = b \cos β t - a \sin β t . \end{aligned}

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Consequently,

\begin{aligned} (\begin{array}{c} u (t) \\ v (t) \end{array}) & = (\begin{array}{c} a \cos β t + b \sin β t \\ b \cos β t - a \sin β t \end{array}) \\ = a (\begin{array}{c} \cos β t \\ - \sin β t \end{array}) + b (\begin{array}{c} \sin β t \\ \cos β t \end{array}) \\ = a x_{Re} (t) + b x_{Im} (t) . \end{aligned}

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Notice that the solutions

x (t) = c_{1} (\begin{matrix} \cos β t \\ - \sin β t \end{matrix}) + c_{2} (\begin{matrix} \sin β t \\ \cos β t \end{matrix})

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are periodic with period

2 π / β .

This type of system is called a center.

Example 5.4.2.

Consider the initial value problem

\begin{aligned} \frac{d x}{d t} & = 2 y \\ \frac{d y}{d t} & = - 2 x \\ x (0) & = 1 \\ y (0) & = 2. \end{aligned}

The eigenvalues of this system are

λ = \pm 2 i .

Therefore, the general solution to the system is

\begin{aligned} x (t) & = c_{1} \cos 2 t + c_{2} \sin 2 t \\ y (t) & = - c_{1} \sin 2 t + c_{2} \cos 2 t . \end{aligned}

Using the initial conditions to solve for

c_{1}

and

c_{2},

the solution to our initial value problem is

\begin{aligned} x (t) & = \cos 2 t + 2 \sin 2 t \\ y (t) & = - \sin 2 t + 2 \cos 2 t . \end{aligned}

The phase portrait is a circle of radius 2 about the origin (Figure 5.4.3).

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Subsection 5.4.2 Spiral Sinks and Sources

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Now let us consider the system

x^{'} = A x,

where

A = (\begin{matrix} α & β \\ - β & α \end{matrix})

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and

α

and

β

are nonzero real numbers. The characteristic equation of

A

λ^{2} - 2 α λ + (α^{2} + β^{2}) = 0,

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so the eigenvalues are

λ = α \pm i β .

We can use the equation

(α - (α + i β)) x + β y = 0

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to show that

(1, i)

is an eigenvector for

α + i β .

Therefore, we have a complex solution of the form

\begin{aligned} x (t) & = e^{(α + i β) t} (\begin{array}{c} 1 \\ i \end{array}) \\ = e^{α t} (\begin{array}{c} \cos β t \\ - \sin β t \end{array}) + i e^{α t} (\begin{array}{c} \sin β t \\ \cos β t \end{array}) \\ = x_{Re} (t) + i x_{Im} (t) . \end{aligned}

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As before, both

x_{Re} (t) = e^{α t} (\begin{matrix} \cos β t \\ - \sin β t \end{matrix}) and x_{Im} (t) = e^{α t} (\begin{matrix} \sin β t \\ \cos β t \end{matrix})

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are real solutions to

x^{'} = A x .

Furthermore, these solutions are linearly independent. Indeed,

x_{Re}

cannot be a multiple of

x_{Im}

for all values of

t .

Thus, we have the general solution

x (t) = c_{1} e^{α t} (\begin{matrix} \cos β t \\ - \sin β t \end{matrix}) + c_{2} e^{α t} (\begin{matrix} \sin β t \\ \cos β t \end{matrix}) .

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The

e^{α t}

factor tells us that the solutions either spiral into the origin if

α < 0

or spiral out to infinity if

α > 0 .

In this case we say that the equilibrium points are spiral sinks and spiral sources, respectively.

Example 5.4.4.

Consider the initial value problem

\begin{aligned} \frac{d x}{d t} & = - x / 10 + y \\ \frac{d y}{d t} & = - x - y / 10 \\ x (0) & = 2 \\ y (0) & = 2. \end{aligned}

The matrix

(\begin{matrix} - 1 / 10 & 1 \\ - 1 & - 1 / 10 \end{matrix})

has eigenvalues

λ = - 1 / 10 \pm i .

The eigenvalue

λ = - 1 / 10 + i

has an eigenvector

v = (1, i) .

The complex solution of our system is

\begin{aligned} x (t) & = e^{(- 1 / 10 + i) t} (\begin{array}{c} 1 \\ i \end{array}) \\ = e^{- t / 10} e^{i t} (\begin{array}{c} 1 \\ i \end{array}) \\ = e^{- t / 10} (\cos t + i \sin t) (\begin{array}{c} 1 \\ i \end{array}) \\ = e^{- t / 10} (\begin{array}{c} \cos t + i \sin t \\ - \sin t + i \cos t \end{array}) \\ = e^{- t / 10} (\begin{array}{c} \cos t \\ - \sin t \end{array}) + i e^{- t / 10} (\begin{array}{c} \sin t \\ \cos t \end{array}) \end{aligned}

The real and imaginary parts of this solution are

x_{Re} (t) = e^{- t / 10} (\begin{matrix} \cos t \\ - \sin t \end{matrix}) and x_{Im} (t) = e^{- t / 10} (\begin{matrix} \sin t \\ \cos t \end{matrix}),

respectively. Thus, we have the general solution

x (t) = c_{1} e^{- t / 10} (\begin{matrix} \cos t \\ - \sin t \end{matrix}) + c_{2} e^{- t / 10} (\begin{matrix} \sin t \\ \cos t \end{matrix}) .

Applying our initial conditions, we can determine that

c_{1} = 2

and

c_{2} = 2;

hence, the solution to our initial value problem is

x (t) = 2 e^{- t / 10} (\begin{matrix} \cos t + \sin t \\ \cos t - \sin t \end{matrix}) .

The phase portrait of this solution indicates that we do indeed have a spiral sink (Figure 5.4.5).

Example 5.4.6.

The initial value problem

\begin{aligned} \frac{d x}{d t} & = x / 10 + y \\ \frac{d y}{d t} & = - x + y / 10 \\ x (0) & = 0 \\ y (0) & = 1 / 2. \end{aligned}

The matrix

(\begin{matrix} 1 / 10 & 1 \\ - 1 & 1 / 10 \end{matrix})

has an eigenvector

(1, - i)

with eigenvalue

λ = 1 / 10 - i .

Thus, the complex solution is

x (t) = e^{(1 / 10 - i) t} (\begin{matrix} 1 \\ - i \end{matrix}) .

Following the procedure that we used in the previous example, the solution to our initial value problem is

x (t) = \frac{1}{2} e^{t / 10} (\begin{matrix} \sin t \\ \cos t \end{matrix}),

and the phase portrait is a spiral source (Figure 5.4.7).

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Activity 5.4.1. Systems with Complex Eigenvalues.

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Consider the system

d x / d t = A x,

where

A = (\begin{matrix} 7 & 4 \\ - 4 & 7 \end{matrix})

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(a)

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Find the eigenvalues,

λ

and

\overset{―}{λ}

A .

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(b)

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Find eigenvectors,

v

and

\overset{―}{v}

for the eigenvalues

λ

and

\overset{―}{λ},

respectively.

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(c)

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Find the complex solution to the system

d x / d t = A x .

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(d)

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Find the real solution to the system

d x / d t = A x .

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(e)

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Is the origin a spiral source or a spiral sink? Sketch a solution curve in the

x y

-plane.

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Subsection 5.4.3 Solving Systems with Complex Eigenvalues

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Suppose that we have the linear system

x^{'} = A x,

where

A = (\begin{matrix} a & b \\ c & d \end{matrix}) .

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The characteristic polynomial of

A

p (λ) = λ^{2} - (a + d) λ + (a d - b c) .

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(a + d)^{2} - 4 (a d - b c) < 0,

then the eigenvalues of

A

are complex, and we cannot apply the strategy that we used to determine the general solution in the case of distinct real roots.

Example 5.4.8.

The system

x^{'} = A x,

where

A = (\begin{matrix} - 1 & - 2 \\ 4 & 3 \end{matrix}) .

The characteristic polynomial of

A

λ^{2} - 2 λ + 5

and so the eigenvalues are complex conjugates,

λ = 1 + 2 i

and

\overset{―}{λ} = 1 - 2 i .

It is easy to show that an eigenvector for

λ = 1 + 2 i

v = (1, - 1 - i) .

Recalling that

e^{i θ} = \cos θ + i \sin θ,

\begin{aligned} x (t) & = e^{(1 + 2 i) t} v \\ = e^{(1 + 2 i) t} (\begin{array}{c} 1 \\ - 1 - i \end{array}) \\ = e^{t} e^{2 i t} (\begin{array}{c} 1 \\ - 1 - i \end{array}) \\ = e^{t} (\cos 2 t + i \sin 2 t) (\begin{array}{c} 1 \\ - 1 - i \end{array}) \\ = e^{t} (\begin{array}{c} \cos 2 t + i \sin 2 t \\ (- 1 - i) (\cos 2 t + i \sin 2 t) \end{array}) \\ = e^{t} (\begin{array}{c} \cos 2 t + i \sin 2 t \\ (- \cos 2 t + \sin 2 t) + i (- \cos 2 t - \sin 2 t) \end{array}) \\ = e^{t} (\begin{array}{c} \cos 2 t \\ - \cos 2 t + \sin 2 t \end{array}) + i e^{t} (\begin{array}{c} \sin 2 t \\ - \cos 2 t - \sin 2 t \end{array}) \end{aligned}

is a complex solution to our system. Taking the real and imaginary parts of this solution, we obtain the general solution to our system

x (t) = c_{1} e^{t} (\begin{matrix} \cos 2 t \\ - \cos 2 t + \sin 2 t \end{matrix}) + c_{2} e^{t} (\begin{matrix} \sin 2 t \\ - \cos 2 t - \sin 2 t \end{matrix}) .

Example 5.4.9.

Solve the following linear system with the given initial conditions, and sketch a phase portrait of the solution.

\begin{aligned} x^{'} & = 3 x - 2 y \\ y^{'} & = 13 x + y \\ x (0) & = 1 \\ y (0) & = 3 \end{aligned}

The characteristic polynomial of the associated matrix is

(3 - λ) (1 - λ) + 26,

which gives eigenvalues

2 \pm 5 i .

Choosing eigenvalue

2 + 5 i

(we could have chosen either one) and subtracting this from the main diagonal of the associated matrix gives

(\begin{matrix} 1 - 5 i & - 2 \\ 13 & - 1 - 5 i \end{matrix}),

so we have eigenvector

(\begin{matrix} 1 + 5 i \\ 13 \end{matrix}) .

This gives a complex solution (ignoring initial conditions)

e^{(2 + 5 i) t} (\begin{matrix} 1 + 5 i \\ 13 \end{matrix})

We can write

e^{2 + 5 i} = e^{2 t} (\cos (5 t) + i \sin (5 t)),

so this solution equals

e^{2 t} (\cos (5 t) + i \sin (5 t)) (\begin{matrix} 1 + 5 i \\ 13 \end{matrix}) = e^{2 t} (\begin{matrix} \cos (5 t) + 5 i \cos (5 t) + i \sin (5 t) - 5 \sin (5 t) \\ 13 \cos (5 t) + 13 i \sin (5 t) \end{matrix})

= e^{2 t} (\begin{matrix} \cos (5 t) - 5 \sin (5 t) \\ 13 \cos (5 t) \end{matrix}) + i \cdot e^{2 t} (\begin{matrix} 5 \cos (5 t) + \sin (5 t) \\ 13 \sin (5 t) \end{matrix})

So the set of all real solutions to the differential system (ignoring initial conditions) is

c_{1} e^{2 t} (\begin{matrix} \cos (5 t) - 5 \sin (5 t) \\ 13 \cos (5 t) \end{matrix}) + c_{2} e^{2 t} (\begin{matrix} 5 \cos (5 t) + \sin (5 t) \\ 13 \sin (5 t) \end{matrix})

Plugging in

t = 0

and setting the top equal to the initial

x

value, the bottom equal to the initial

y

value, we get the system

\begin{aligned} c_{1} + 5 c_{2} & = 1 \\ 13 c_{1} & = 3 \end{aligned}

This gives the solutions

c_{1} = \frac{3}{13}, c_{2} = \frac{2}{13};

plugging these into the general equation gives the solution

e^{2 t} (\begin{matrix} \cos (5 t) - \sin (5 t) \\ 3 \cos (5 t) + 2 \sin (5 t) \end{matrix})

Now we are asked to draw a sketch of the solution. Since the real part of the eigenvalue (

2

) is positive, we have a spiral source. We also have to determine whether the spiral goes clockwise or counter-clockwise as we spiral out. The matrix associated to the equation is

A = (\begin{matrix} 3 & - 2 \\ 13 & 1 \end{matrix}) .

If we multiply this matrix by

(\begin{matrix} 1 \\ 0 \end{matrix})

we get

(\begin{matrix} 3 & - 2 \\ 13 & 1 \end{matrix}) \cdot (\begin{matrix} 1 \\ 0 \end{matrix})

= (\begin{matrix} 3 \\ 13 \end{matrix}) .

This means that at the point

(1, 0)

a solution curve will be pointing toward the diretion

(\begin{matrix} 3 \\ 13 \end{matrix}),

so we are spiraling counter-clockwise away from the origin. So we must draw a curve passing through

(\begin{matrix} 1 \\ 3 \end{matrix})

spiraling counter-clockwise outward.

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The nature of the equilibrium solution is determined by the factor

e^{α t}

in the solution. If

α < 0,

the equilibrium point is a spiral sink. If

α > 0,

the equilibrium point is a spiral source. If

α = 0,

the equilibrium point is a center.

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Although we have outlined a procedure to find the general solution of

x^{'} = A x

A

has complex eigenvalues, we have not shown that this method will work in all cases. We will do so in Section 6.1.

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Activity 5.4.2. Planar Systems with Complex Eigenvalues.

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Consider the system

d x / d t = A x,

where

A = (\begin{matrix} 7 & - 4 \\ 10 & - 5 \end{matrix})

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(a)

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Find the eigenvalues,

λ

and

\overset{―}{λ}

A .

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(b)

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Find eigenvectors,

v

and

\overset{―}{v}

for the eigenvalues

λ

and

\overset{―}{λ},

respectively.

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(c)

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Find the complex solution to the system

d x / d t = A x .

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(d)

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Find the real solution to the system

d x / d t = A x .

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(e)

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Is the origin a spiral source or a spiral sink? Sketch a solution curve in the

x y

-plane.

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Subsection 5.4.4 Important Lessons

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If

$A = (\begin{matrix} α & β \\ - β & α \end{matrix}),$

then $A$ has two complex eigenvalues, $λ = α \pm i β .$ The general solution to the system $x^{'} = A x$ is

$x (t) = c_{1} e^{α t} (\begin{matrix} \cos β t \\ - \sin β t \end{matrix}) + c_{2} e^{α t} (\begin{matrix} \sin β t \\ \cos β t \end{matrix}) .$

If $α < 0,$ the equilibrium point is a spiral sink. If $α > 0,$ the equilibrium point is a spiral source.