ODE-ProjectODE-Project Hints and Answers to Selected Exercises

Appendix B 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

1.4.6.5.

Hint.

This equation is a first-order linear equation (Section 1.5), but it is possible to find the analytic solution using Sage (Subsection 1.2.10).

1.4.6.8.

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.

1.5.8.25. Exact Differential Equations.

Hint.

For (e), rewrite the equation as

(y d x + x d y) + (- x y^{2} d x + x^{2} y^{2} d y) = 0

and consider the integrating factor

μ (x, y) = 1 / (x y)^{2} .

1.5.8.28.

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

1.6.5.1.

Hint.

There exists a unique solution to $y^{'} = y^{2} + y^{3},$ $y (0) = 1,$ since $f (t, y) = y^{2} + y^{3}$ and $\partial f (t, y) / \partial y = 2 y + 3 y^{2}$ are continuous at the point $(0, 1) .$
The Existence and Uniqueness Theorem does not apply to $y^{'} = \sqrt[4]{y},$ $y (1) = 0,$ since $f (t, y) = \sqrt[4]{y}$ is not continuous at $(1, 0) .$
There exists a unique solution to $y^{'} = \sqrt[4]{y},$ $y (1) = 1,$ since $f (t, y) = \sqrt[4]{y}$ and $\partial f (t, y) / \partial y = y^{- 3 / 4} / 4$ are both continuous at the point $(1, 1) .$
The Existence and Uniqueness Theorem does not apply to $x^{'} = t / (x^{2} - 4),$ $x (0) = 2,$ since $f (t, x) = t / (x^{2} - 4)$ is not continuous at $(0, 2) .$
There exists a unique solution to $x^{'} = t / (x^{2} - 4),$ $x (2) = 0,$ since $f (t, x) = t / (x^{2} - 4)$ and $\partial f (t, x) / \partial x = - 2 t x / (x^{2} - 4)^{2}$ are both continuous at the point $(2, 0) .$
There exists a unique solution to $y^{'} = x \sin y,$ $y (0) = 0,$ since $f (x, y) = x \sin y$ and $\partial f (x, y) / \partial y = x \cos y$ are both continuous at the point $(0, 0) .$
The Existence and Uniqueness Theorem does not apply to $y^{'} = 1 / (t - 1) y + 2 t,$ $y (1) = 1,$ since $f (t, y) = 1 / (t - 1) y + 2 t$ is not continuous at $(1, 1) .$

1.6.5.3.

Hint.

(b) Make sure that the derivative of

y (t)

exists at

t = t_{0} .

3 Linear Systems
3.2 Planar Systems
3.2.6 Exercises

3.2.6.10.

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.

4 Second-Order Linear Equations
4.1 Homogeneous Linear Equations
4.1.6 Exercises

4.1.6.31.

Hint.

Pay careful attention to units.

4.1.6.32.

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, $v (t) = c_{1} + c_{2} t .$

4.1.6.33.

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

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

4.1.6.38. Euler Equations.

Hint.

Show that

\begin{aligned} \frac{d y}{d t} & = \frac{d x}{d t} \frac{d y}{d x} \\ \frac{d^{2} y}{d t^{2}} & = {(\frac{d x}{d t})}^{2} \frac{d^{2} y}{d x^{2}} + \frac{d^{2} x}{d t^{2}} \frac{d y}{d x} . \end{aligned}

4.1.6.39. Higher Order Linear Equqtions with Constant Coefficients.

Hint.

Show that

\begin{aligned} \frac{d y}{d t} & = \frac{d x}{d t} \frac{d y}{d x} \\ \frac{d^{2} y}{d t^{2}} & = {(\frac{d x}{d t})}^{2} \frac{d^{2} y}{d x^{2}} + \frac{d^{2} x}{d t^{2}} \frac{d y}{d x} . \end{aligned}

4.2 Forcing
4.2.7 Exercises

4.2.7.25.

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) .

W (f, g) (t) = det (\begin{matrix} f (t) & g (t) \\ f^{'} (t) & g^{'} (t) \end{matrix}) = 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) (t) = det (\begin{matrix} f (t) & g (t) \\ f^{'} (t) & g^{'} (t) \end{matrix}) = 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.

4.2.7.26.

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
$W [y_{1}, y_{2}] (t) = c \exp (- \int \frac{3}{2 t} d t) = c \exp (- \frac{3}{2} \ln t) = c t^{- 3 / 2} .$
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
$\begin{matrix} (4.2.5) & (y_{1} y_{2}^{″} - y_{1}^{″} y_{2}) + p (t) (y_{1} y_{2}^{'} - y_{1}^{'} y_{2}) = 0. \end{matrix}$
If
$W (t) = W (y_{1}, y_{2}) (t) = y_{1} y_{2}^{'} - y_{1}^{'} y_{2},$
then
$W^{'} = y_{1} y_{2}^{″} - y_{1}^{″} y_{2},$
and equation (4.2.5) becomes
$W^{'} + p (t) W = 0.$
This equation is separable with solution
$W (t) = c \exp (- \int p (t) d t) .$

4.2.7.27.

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 (4.2.6), 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
$\begin{aligned} y & = y_{h} + y_{p} \\ = c_{1} \cos 2 t + c_{2} \sin 2 t - 3 \sin t \cos 2 t + [\frac{3}{2} \ln | \csc t - \cot t | + 3 \cos t] \sin 2 t . \end{aligned}$

4.3 Sinusoidal Forcing
4.3.5 Exercises

4.3.5.2.

Hint.

Assume the complex solution has form

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

4.3.5.12.

Hint.

Assume the complex solution has form

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

5 Nonlinear Systems
5.3 More Nonlinear Mechanics
5.3.6 Exercises

5.3.6.17. Lobster Navigation.

Hint.

You may find Sage useful.

🔗

Prev Top Next

Appendix B 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

1.4.6.5.

1.4.6.8.

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.

1.5.8.25. Exact Differential Equations.

1.5.8.28.

1.6 Existence and Uniqueness of Solutions1.6.5 Exercises

1.6.5.1.

1.6.5.3.

3 Linear Systems3.2 Planar Systems3.2.6 Exercises

3.2.6.10.

4 Second-Order Linear Equations4.1 Homogeneous Linear Equations4.1.6 Exercises

4.1.6.31.

4.1.6.32.

4.1.6.33.

4.1.6.38. Euler Equations.

4.1.6.39. Higher Order Linear Equqtions with Constant Coefficients.

4.2 Forcing4.2.7 Exercises

4.2.7.25.

4.2.7.26.

4.2.7.27.

4.3 Sinusoidal Forcing4.3.5 Exercises

4.3.5.2.

4.3.5.12.

5 Nonlinear Systems5.3 More Nonlinear Mechanics5.3.6 Exercises

5.3.6.17. Lobster Navigation.

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

3 Linear Systems
3.2 Planar Systems
3.2.6 Exercises

4 Second-Order Linear Equations
4.1 Homogeneous Linear Equations
4.1.6 Exercises

4.2 Forcing
4.2.7 Exercises

4.3 Sinusoidal Forcing
4.3.5 Exercises

5 Nonlinear Systems
5.3 More Nonlinear Mechanics
5.3.6 Exercises