ODE-Project Hints and Answers to Selected Exercises

Appendix C Hints and Answers to Selected Exercises

1 A First Look at Differential Equations
1.1 Modeling with Differential Equations
1.1.9 Exercises

1.1.9.29.

Hint.

Rewriting the differential equation as

x^{'} + a x - q (t) = 0

and using the fact that

x^{'} (t) = - a C e^{- a t} - a e^{- a t} \int_{t_{0}}^{t} e^{a s} q (s) d s + b (t),

we see that

\begin{aligned} x^{'} (t) + a x (t) - q (t) & = - a C e^{- a t} - a e^{- a t} \int_{t_{0}}^{t} e^{a s} q (s) d s + q (t) \\ + a C e^{- a t} + a e^{- a t} \int_{t_{0}}^{t} e^{a s} q (s) d s - q (t) \\ = 0. \end{aligned}

1.1.9.31.

Hint.

Think about the limit of the interaction term as the number of prey becomes very large.

1.4 Analyzing Equations Numerically
1.4.6 Exercises

1.4.6.7.

Hint.

Hints for part (2):

For fixed $i$ show that

$\begin{aligned} a_{i + 1} & \leq (1 + s) a_{i} + t \\ \leq (1 + s) [(1 + s) a_{i - 1} + t] + t \\ \leq (1 + s) {(1 + s) [(1 + s) a_{i - 2} + t] + t} + t \\ ⋮ \\ \leq (1 + s)^{i + 1} a_{0} + [1 + (1 + s) + (1 + s)^{2} + \dots + (1 + s)^{i}] t . \end{aligned}$
Now use a geometric sum to show that

$a_{i + 1} \leq (1 + s)^{i + 1} a_{0} + \frac{t}{s} [(1 + s)^{i + 1} - 1] = (1 + s)^{i + 1} (\frac{t}{s} + a_{0}) - \frac{t}{s} .$
Apply part (1) to derive

$a_{i + 1} \leq e^{(i + 1) s} (\frac{t}{s} + a_{0}) - \frac{t}{s} .$

1.5 First-Order Linear Equations
1.5.8 Exercises

1.5.8.16.

Hint.

y = \frac{3 x^{4} + 8 x^{3} + 6 x^{2} + 12}{6 (x + 1)^{2}}

1.5.8.17.

Hint.

y = (e^{x} + 1) / (x + 1)^{2}

1.5.8.19.

Hint.

y = e^{x} \sec x

1.5.8.20.

Hint.

y = \sin x / (x + 2)

1.5.8.21.

Hint.

x (t)

is the amount of salt in the tank at time

t,

we know that

x (0) = 10 .

The volume of the tank is

V = 200 + 5 t .

We can model the amount of salt in the tank at time

t

with a differential equation,

\begin{aligned} \frac{d x}{d t} & = rate in - rate out \\ = 10 (0.1) - 5 \frac{x}{V} \\ = 1 - 5 \frac{x}{200 + 5 t} \\ = 1 - \frac{x}{40 + t} . \end{aligned}

The resulting equation

\frac{d x}{d t} + \frac{1}{40 + t} x = 1

is a first order linear differential equation. An integrating factor for this equation is given by

μ (t) = \exp (\int \frac{1}{40 + t} d t) = 40 + t .

Multiplying both sides of the differential equation by

μ (t),

we have

\frac{d}{d t} [(40 + t) x] = (40 + t) \frac{d x}{d t} + x = (40 + t) (\frac{d x}{d t} + \frac{1}{40 + t} x) = 40 + t .

Integrating both sides of this equation, we obtain

(40 + t) x = 40 t + \frac{t^{2}}{2} + C .

Using the intial condition

x (0) = 10,

we can determine that

C = 400

x (t) = \frac{t^{2} + 80 t + 800}{2 t + 80} .

The tank is full at time

t = 400 / 5 = 80,

and the tank contains

x (80) = 170 / 3 \approx 56.67

kilograms of salt when the tank is full.

1.5.8.26.

Hint.

If $y = y_{1} + 1 / v,$ then $y^{'} = y_{1}^{'} - v^{'} / v^{2} .$ Substituting into our original equation, we obtain

$y^{'} = y_{1}^{'} - \frac{v^{'}}{v^{2}} = p + q y_{1} + r y_{1}^{2} - \frac{v^{'}}{v^{2}} .$

On the other hand,

$\begin{aligned} y^{'} & = p + q (y_{1} + \frac{1}{v}) + r {(y_{1} + \frac{1}{v})}^{2} \\ = p + q y_{1} + \frac{q}{v} + r y_{1}^{2} + \frac{2 r y_{1}}{v} + \frac{r}{v^{2}} \\ = y_{1}^{'} + \frac{q}{v} + \frac{2 r y_{1}}{v} + \frac{r}{v^{2}} . \end{aligned}$

Therefore,

$- \frac{v^{'}}{v^{2}} = \frac{q}{v} + \frac{2 r y_{1}}{v} + \frac{r}{v^{2}},$

which is just the first-order linear equation

$v^{'} + [q (t) + 2 r (t) y_{1} (t)] v = - r (t) .$
$y = t + \frac{1}{C - t}$
$y (t) = \frac{1}{C \cos t - \sin t} + \sin t$
$y (t) = 2 + \frac{1}{C e^{t} - 1}$

1.6 Existence and Uniqueness of Solutions
1.6.5 Exercises

1.6.5.2.

Hint.

(b) Make sure that the derivative of

y (t)

exists at

t = t_{0} .

2 Second-Order Linear Equations
2.1 Homogeneous Linear Equations
2.1.7 Exercises

2.1.7.31.

Hint.

Pay careful attention to units.

2.1.7.32.

Hint.

Observe that

$\begin{aligned} a x_{1}^{″} + b_{1}^{'} + c x_{1} & = a {(\frac{- b}{2 a})}^{2} e^{- b t / 2 a} + b (\frac{- b}{2 a}) e^{- b t / 2 a} + c e^{- b t / 2 a} \\ = e^{- b t / 2 a} (\frac{b^{2}}{4 a} - \frac{b^{2}}{2 a} + c) \\ = e^{- b t / 2 a} (\frac{- b^{2} + 4 a c}{4 a}) \\ = 0. \end{aligned}$
If $y = v (t) x_{1} (t) = v (t) e^{- b t / 2 a}$ is a solution to our differential equation, then

$\begin{aligned} a y^{″} + b y^{'} + c y & = a (v^{″} x_{1} + 2 v^{'} x_{1}^{'} + v x_{1}^{″}) + b (v^{'} x_{1} + v x_{1}^{'}) + c v x_{1} \\ = a v^{″} x_{1} + 2 a v^{'} x_{1}^{'} + b v^{'} x_{1} + v (a x_{1}^{″} + b x_{1}^{'} + c x_{1}) \\ = a v^{″} e^{- b t / 2 a} + [2 a (\frac{- b}{2 a}) e^{- b t / 2 a} + b e^{- b t / 2 a}] v^{'} \\ = a v^{″} e^{- b t / 2 a} \\ = 0. \end{aligned}$

Since $a \neq 0,$ we know that $v^{″} = 0 .$ Hence, $v (t) = c_{1} + c_{2} t .$

2.1.7.33.

Hint.

$\begin{aligned} x^{″} + p x^{'} + q x & = (v^{″} x_{1} + 2 v^{'} x_{1}^{'} + v x_{1}^{″}) + p (v^{'} x_{1} + v x_{1}^{'}) + q (v x_{1}) \\ = x_{1} v^{″} + 2 v^{'} x_{1}^{'} + p x_{1} v^{'} + v (x_{1}^{″} + p x_{1}^{'} + q x_{1}) \\ = x_{1} v^{″} + (2 x_{1}^{'} + p x_{1}) v^{'} \\ = 0. \end{aligned}$
If $u = v^{'},$ then $x_{1} u^{'} + (2 x_{1}^{'} + p x_{1}) u = 0 .$
If $x_{1} (t) = 1 / t,$ then

$2 t^{2} x_{1}^{″} + 3 t x_{1}^{'} - x_{1} = 2 t^{2} (\frac{2}{t^{3}}) + 3 t (\frac{- 1}{t^{2}}) - \frac{1}{t} = 0.$

If we assume that $x = v / t$ is a second solution, then

$2 t^{2} x^{″} + 3 t x^{'} - x = 2 t v^{″} - v^{'} = 0.$

If we let $u = v^{'},$ then a solution of $2 t u^{'} - u = 0$ is $u = \sqrt{t}$ and $v = \int \sqrt{t} d t = 2 t^{3 / 2} / 3 .$ Therefore, the second solution to our equation is

$x = \frac{v}{t} = \frac{2}{3} \sqrt{t} .$

2.2 Forcing
2.2.7 Exercises

2.2.7.2.

Answer.

The solution to the associated homogeneous equation is

c_{1} e^{2 x} + c_{2} e^{- x},

and we do not see any

e^{2 x}

e^{- x}

in the right-hand-side, so we look at Table 2.2.12 and see the solution has the form

y (x) = A x^{2} + B x + C .

We then get

y^{'} (x) = 2 A x + B, y^{″} (x) = 2 A,

y^{″} - y^{'} - 2 y = 2 A - (2 A x + B) - 2 (A x^{2} + B x + C) =

- 2 A x^{2} + (- 2 A - 2 B) x + (2 A - B - 2 C) = 4 x^{2} - 2 x - 9

We then have to equate the coefficients.

- 2 A = 4

gives

A = - 2,

- 2 A - 2 B = - 2

gives

B = - A + 1 = 3,

and

2 A - B - 2 C = - 9

gives

C = 1,

so the answer is

y (x) = - 2 x^{2} + 3 x + 1

2.2.7.25.

Hint.

Suppose that that

f (t)

and

g (t)

are linearly dependent on an interval

I = (a, b) .

Then one function is a multiple of the other, say

f (t) = c g (t) .

Thus,

f^{'} (t) = c g^{'} (t)

and

W (f, g) = f (t) g^{'} (t) - f^{'} (t) g (t) = c g (t) g^{'} (t) - c g^{'} (t) g (t) = 0.

Conversely, suppose that

W (f, g) = 0

for all

t

(a, b) .

g = 0,

then

0 f = g

and the two functions are linearly dependent. Assume that

g (t_{0}) \neq 0

for some

t_{0}

(a, b) .

Since

g

is differentiable, it must also be continuous and there is some interval

(c, d)

contained in

(a, b)

such that

t_{0} \in (c, d)

and

g

does not vanish on this interval. Therefore,

\frac{d}{d t} (\frac{f}{g}) = \frac{f^{'} g - f g^{'}}{g^{2}} = - \frac{W (f, g)}{g^{2}} = 0,

and

f / g

is constant on the interval

(c, d) .

Thus,

f (t_{0}) = c g (t_{0})

and

f^{'} (t_{0}) = c g^{'} (t_{0}) .

Since

f

and

c g

are both solutions to the differential equation

y^{″} + p y^{'} + q y = 0

and have the same initial condition,

f (t) = c g (t)

for all

t \in (a, b)

by the existence and uniqueness theorem. Consequently,

f

and

g

are linearly dependent.

2.2.7.26.

Hint.

We can rewrite $2 t^{2} y^{″} + 3 t y^{'} - y = 0$ as
$y^{″} + \frac{3}{2 t} y^{'} - \frac{1}{2 t^{2}} y = 0.$
Since $p (t) = 1 / 2 t,$ Abel’s Theorem tells us that
$W (y_{1}, y_{2}) = c \exp (- \int \frac{3}{2 t} d t) = c \exp (- \frac{3}{2} \ln t) = c t^{- 3 / 2} .$
Since $y_{1}$ and $y_{2}$ are solutions to our differential equation, we know that
$\begin{matrix} y_{1}^{″} + p (t) y_{1}^{'} + q (t) y_{1} = 0 \\ y_{2}^{″} + p (t) y_{2}^{'} + q (t) y_{2} = 0. \end{matrix}$
Multiplying the first equation by $y_{2}$ and the second equation by $y_{1}$ and subtracting, we obtain
$\begin{matrix} (2.2.5) & (y_{1} y_{2}^{″} - y_{1}^{″} y_{2}) + p (t) (y_{1} y_{2}^{'} - y_{1}^{'} y_{2}) = 0. \end{matrix}$
If
$W (t) = W (y_{1}, y_{2}) = y_{1} y_{2}^{'} - y_{1}^{'} y_{2},$
then
$W^{'} = y_{1} y_{2}^{″} - y_{1}^{″} y_{2},$
and equation (2.2.5) becomes
$W^{'} + p (t) W = 0.$
This equation is separable with solution
$W (t) = c \exp (- \int p (t) d t) .$

2.2.7.27.

Hint.

If $y_{p} = u_{1} y_{1} + u_{2} y_{2},$ then
$\begin{matrix} y_{p}^{'} = u_{1}^{'} y_{1} + u_{1} y_{1}^{'} + u_{2}^{'} y_{2} + u_{2} y_{2}^{'} = u_{1} y_{1}^{'} + u_{2} y_{2}^{'} \\ y_{p}^{″} = u_{1}^{'} y_{1}^{'} + u_{1} y_{1}^{″} + u_{2}^{'} y_{2}^{'} + u_{2} y_{2}^{″} . \end{matrix}$
Substituting these expressions into equation (2.2.6), we have
$\begin{aligned} y_{p}^{″} + p y_{p}^{'} + q y_{p} & = (u_{1}^{'} y_{1}^{'} + u_{1} y_{1}^{″} + u_{2}^{'} y_{2}^{'} + u_{2} y_{2}^{″}) + p (u_{1} y_{1}^{'} + u_{2} y_{2}^{'}) \\ + q (u_{1} y_{1} + u_{2} y_{2}) \\ = u_{1} [y_{1}^{″} + p y_{1}^{'} + q y_{1}] + u_{2} [y_{2}^{″} + p y_{2}^{'} + q y_{2}] + u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'} \\ = u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'} \\ = f (t) . \end{aligned}$
If we solve the system
$\begin{aligned} u_{1}^{'} (t) y_{1} (t) + u_{2}^{'} (t) y_{2} (t) & = 0 \\ u_{1}^{'} (t) y_{1}^{'} (t) + u_{2}^{'} (t) y_{2}^{'} (t) & = f (t) . \end{aligned}$
for $u_{1}^{'}$ and $u_{2}^{'},$ we obtain
$\begin{aligned} u_{1}^{'} (t) & = \frac{- y_{2} (t) f (t)}{W [y_{1}, y_{2}] (t)} \\ u_{2}^{'} (t) & = \frac{y_{1} (t) f (t)}{W [y_{1}, y_{2}] (t)} . \end{aligned}$
Integrate the two equations from part (2).
The general solution to the homogeneous equation $y^{″} + 4 y = 0$ is
$y_{h} = c_{1} \cos 2 t + c_{2} \sin 2 t .$
To find a particular solution, assume that the solution has the form
$y_{p} = u_{1} (t) \cos 2 t + u_{2} (t) \sin 2 t .$
By part (2)
$\begin{aligned} u_{1}^{'} (t) & = - 3 \cos t \\ u_{2}^{'} (t) & = \frac{3}{2} \csc t - 3 \sin t . \end{aligned}$
Integrating, we obtain
$\begin{aligned} u_{1} (t) & = - 3 \sin t \\ u_{2} (t) & = \frac{3}{2} \ln | \csc t - \cot t | + 3 \cos t . \end{aligned}$
Therefore,
$\begin{aligned} y_{p} (t) & = u_{1} (t) y_{1} (t) + u_{2} (t) y_{2} (t) \\ = - 3 \sin t \cos 2 t + [\frac{3}{2} \ln | \csc t - \cot t | + 3 \cos t] \sin 2 t, \end{aligned}$
and the general solution is
$\begin{aligned} y & = y_{h} + y_{p} \\ = c_{1} \cos 2 t + c_{2} \sin 2 t - 3 \sin t \cos 2 t + [\frac{3}{2} \ln | \csc t - \cot t | + 3 \cos t] \sin 2 t . \end{aligned}$

2.3 Sinusoidal Forcing
2.3.5 Exercises

2.3.5.2.

Hint.

Assume the complex solution has form

y_{c} = A e^{3 i t} .

2.3.5.12.

Hint.

Assume the complex solution has form

y_{c} = A e^{3 i t} .

3 The Laplace Transform
3.1 The Laplace Transform
3.1.7 Exercises

3.1.7.15.

Answer.

If the function in question were simply

\frac{1}{s^{4}},

we would observe that

L (t^{3}) (s) = \frac{6}{s^{4}},

so the answer would be

\frac{t^{3}}{6} .

Since we subtract 2 from the argument

s,

we multiply by

e^{2 t},

so the answer is

\frac{e^{2 t} t^{3}}{6} .

3.1.7.16.

Answer.

We will split this into partial fractions via

\frac{1}{(s + 1)^{2} (s^{2} - 4)} = \frac{A (s + 1) + B}{(s + 1)^{2}} + \frac{C}{s + 2} + \frac{D}{s - 2}

Clearing denominators gives

1 = [A (s + 1) + B] (s^{2} - 4) + C (s - 2) (s + 1)^{2} + D (s + 2) (s + 1)^{2}

Plugging in

s = - 1

yields

- 3 B = 1,

B = \frac{- 1}{3} .

Plugging in

s = 2

yields

36 D = 1,

D = \frac{1}{36} .

Plugging in

s = - 2

yields

- 4 C = 1,

C = \frac{- 1}{4} .

Plugging in

s = 0

and using our values for

B, C,

and

D

yields

- 4 A + \frac{17}{9} = 1,

A = \frac{2}{9}

So we are to find the inverse Laplace of

F (s) = \frac{\frac{2}{9} (s + 1) - \frac{1}{3}}{(s + 1)^{2}} - \frac{1 / 4}{s + 2} + \frac{1 / 36}{s - 2}

= \frac{2 / 9}{s + 1} - \frac{1 / 3}{(s + 1)^{2}} - \frac{1 / 4}{s + 2} + \frac{1 / 36}{s - 2}

This gives us

f (t) = \frac{2}{9} e^{- t} - \frac{1}{3} t e^{- t} - \frac{1}{4} e^{- 2 t} + \frac{1}{36} e^{2 t}

3.2 Solving Initial Value Problems
3.2.6 Exercises

3.2.6.1.

Answer.

Taking the Laplace transformation of both sides yields

s^{2} Y (s) - s - 2 s Y (s) + 2 - 3 Y (s) = \frac{24}{s + 3}

Y (s) = \frac{24}{(s - 3) (s + 1) (s + 3)} + \frac{s - 2}{(s - 3) (s + 1)}

A partial fractions decomposition reveals

Y (s) = \frac{1}{s - 3} + \frac{- 3}{s + 1} + \frac{2}{s + 3} + \frac{1 / 4}{s - 3} + \frac{3 / 4}{s + 1}

So the answer is

y (t) = e^{3 t} - 3 e^{- t} + 2 e^{- 3 t} + \frac{1}{4} e^{3 t} + \frac{3}{4} e^{- t}

y (t) = \frac{5}{4} e^{3 t} - \frac{9}{4} e^{- t} + 2 e^{- 3 t}

3.4 Convolution
3.4.5 Exercises

3.4.5.13.

Answer.

Taking the Laplace transformation of both sides yields

2 s^{2} Y (s) + 5 s Y (s) + 2 Y (s) = G (s)

Therefore,

Y (s) = \frac{G (s)}{2 s^{2} + 5 s + 2} .

We now want to factor

2 s^{2} + 5 s + 2 .

We multiply the

s^{2}

and constant coefficients to get 4, and search for two numbers multiplying to 4 and adding to the

s

coefficient (

5

). We see that

4

and

1

work, and we can write

2 s^{2} + 5 s + 2 = 2 s^{2} + 4 s + s + 2 = 2 s (s + 2) + (s + 2) =

(2 s + 1) (s + 2),

so we get

y (t) = L^{- 1} {\frac{G (s)}{(2 s + 1) (s + 2)}} = g (t) * L^{- 1} {\frac{1}{(2 s + 1) (s + 2)}}

We could use partial fractions or another convolution to calculate

L^{- 1} {\frac{1}{(2 s + 1) (s + 2)}} .

If we choose convolution, we see

L^{- 1} {\frac{1}{(2 s + 1) (s + 2)}}

= \frac{1}{2} e^{- t / 2} * e^{- 2 t}

= \frac{1}{2} \int_{0}^{t} e^{- τ / 2} e^{- 2 (t - τ)} d τ

= \frac{1}{3} (e^{- t / 2} - e^{- 2 t}),

so the final answer is

\frac{1}{3} g (t) * (e^{- t / 2} - e^{- 2 t}) = \frac{1}{3} \int_{0}^{t} g (τ) (e^{\frac{τ - t}{2}} - e^{2 (τ - t)}) d τ

5 Linear Systems and Linearization
5.1 Linear Algebra in a Nutshell
5.1.6 Exercises

5.1.6.4.

Answer.

d e t (A) = 1 (3) - 4 (2) = - 5

5.1.6.6.

Answer.

d e t (A) = 2 (4) - (- 6) (- 2) = - 4

5.2 Planar Systems
5.2.6 Exercises

5.2.6.2.

Answer.

This is equivalent to

x^{'} = A x,

where

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

The characteristic polynomial is

(- 1 - λ) (2 - λ) - 2 (2)

= λ^{2} - λ - 6,

with zeros

λ = - 2, 3 .

We then get

A - (- 2) I = (\begin{matrix} - 1 + 2 & 2 \\ 2 & 2 + 2 \end{matrix}) = (\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix})

λ = - 2

has corresponding eigenvector

(\begin{matrix} - 2 \\ 1 \end{matrix}) .

Analogously,

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

λ = 3

has corresponding eigenvector

(\begin{matrix} 1 \\ 2 \end{matrix}) .

Therefore the set of solutions is

c_{1} e^{- 2 t} (\begin{matrix} - 2 \\ 1 \end{matrix}) + c_{2} e^{3 t} (\begin{matrix} 1 \\ 2 \end{matrix})

5.2.6.6.

Answer.

From exercise 2, we see the general solution to the system

x^{'} = - x + 2 y, y^{'} = 2 x + 2 y

c_{1} e^{- 2 t} (\begin{matrix} - 2 \\ 1 \end{matrix}) + c_{2} e^{3 t} (\begin{matrix} 1 \\ 2 \end{matrix})

Plugging in

t = 0

and setting the top to

19,

the bottom to

8,

we end up at the system

\begin{matrix} - 2 c_{1} + c_{2} = 19 \\ c_{1} + 2 c_{2} = 8 \end{matrix}

Solving this yields

c_{1} = - 6, c_{2} = 7,

and we thus have the solution

- 6 e^{- 2 t} (\begin{matrix} - 2 \\ 1 \end{matrix}) + 7 e^{3 t} (\begin{matrix} 1 \\ 2 \end{matrix})

which can also be written as

\begin{aligned} x (t) = & 12 e^{- 2 t} + 7 e^{3 t} \\ y (t) = & - 6 e^{- 2 t} + 14 e^{3 t} \end{aligned}

5.2.6.10.

Hint.

Assume that your solution must be of the form

x_{p} = (\begin{matrix} a_{2} t^{2} + a_{1} t + a_{0} \\ b_{2} t^{2} + b_{1} t + b_{0} . \end{matrix})

This is called the method of undetermined coefficients.

5.3 Phase Plane Analysis of Linear Systems
5.3.6 Exercises

5.3.6.2.

Answer.

(a) The characteristic polynomial is

(4 - λ) (- 3 - λ) + 6,

with zeros (and thus eigenvalues)

λ = - 2, 3

(b) If the eigenvalues have the same sign, the one with a bigger absolute value is dominant. Since these eigenvalues have different signs, there is no dominant eigenvalue.

λ = - 2

we subtract

- 2

from the main diagonal of the matrix to obtain

A + 2 I = (\begin{matrix} 6 & - 2 \\ 3 & - 1 \end{matrix}),

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

is an eigenvector.

For

λ = 3,

we subtract

3

from the main diagonal to obtain

A - 3 I = (\begin{matrix} 1 & - 2 \\ 3 & - 6 \end{matrix}),

v = (\begin{matrix} 2 \\ 1 \end{matrix})

(d) There are two straight-line solutions, one corresponding to each eigenvalue. For

λ = - 2

we got

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

so the solution set here is the set of multiples of

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

or equivalently, the line

1 y = 3 x .

For

λ = 3

we got the eigenvalue

(\begin{matrix} 2 \\ 1 \end{matrix}),

so the straight-line solution is the line passing through

(\begin{matrix} 2 \\ 1 \end{matrix})

and the origin, or the line

2 y = 1 x .

(e) Since there is one positive and one negative eigenvalue, the equilibrium solution is a saddle.

(f)

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Appendix C Hints and Answers to Selected Exercises

1 A First Look at Differential Equations1.1 Modeling with Differential Equations1.1.9 Exercises

1.1.9.29.

1.1.9.31.

1.4 Analyzing Equations Numerically1.4.6 Exercises

1.4.6.7.

1.5 First-Order Linear Equations1.5.8 Exercises

1.5.8.16.

1.5.8.17.

1.5.8.19.

1.5.8.20.

1.5.8.21.

1.5.8.26.

1.6 Existence and Uniqueness of Solutions1.6.5 Exercises

1.6.5.2.

2 Second-Order Linear Equations2.1 Homogeneous Linear Equations2.1.7 Exercises

2.1.7.31.

2.1.7.32.

2.1.7.33.

2.2 Forcing2.2.7 Exercises

2.2.7.2.

2.2.7.25.

2.2.7.26.

2.2.7.27.

2.3 Sinusoidal Forcing2.3.5 Exercises

2.3.5.2.

2.3.5.12.

3 The Laplace Transform3.1 The Laplace Transform3.1.7 Exercises

3.1.7.15.

3.1.7.16.

3.2 Solving Initial Value Problems3.2.6 Exercises

3.2.6.1.

3.4 Convolution3.4.5 Exercises

3.4.5.13.

5 Linear Systems and Linearization5.1 Linear Algebra in a Nutshell5.1.6 Exercises

5.1.6.4.

5.1.6.6.

5.2 Planar Systems5.2.6 Exercises

5.2.6.2.

5.2.6.6.

5.2.6.10.

5.3 Phase Plane Analysis of Linear Systems5.3.6 Exercises

5.3.6.2.

1 A First Look at Differential Equations
1.1 Modeling with Differential Equations
1.1.9 Exercises

1.4 Analyzing Equations Numerically
1.4.6 Exercises

1.5 First-Order Linear Equations
1.5.8 Exercises

1.6 Existence and Uniqueness of Solutions
1.6.5 Exercises

2 Second-Order Linear Equations
2.1 Homogeneous Linear Equations
2.1.7 Exercises

2.2 Forcing
2.2.7 Exercises

2.3 Sinusoidal Forcing
2.3.5 Exercises

3 The Laplace Transform
3.1 The Laplace Transform
3.1.7 Exercises

3.2 Solving Initial Value Problems
3.2.6 Exercises

3.4 Convolution
3.4.5 Exercises

5 Linear Systems and Linearization
5.1 Linear Algebra in a Nutshell
5.1.6 Exercises

5.2 Planar Systems
5.2.6 Exercises

5.3 Phase Plane Analysis of Linear Systems
5.3.6 Exercises