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Q20E

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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 131
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

Falling Body. In Example 1 of Section 3.4, page 110, we modeled the velocity of a falling body by the initial value problem \({\bf{m}}\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = mg - bv,v(0) = }}{{\bf{v}}_{\bf{o}}}{\bf{ = 0}}\)under the assumption that the force due to air resistance is –bv. However, in certain cases the force due to air resistance behaves more like\({\bf{ - b}}{{\bf{v}}^{\bf{r}}}\), where \({\bf{(r > 1)}}\) is some constant. This leads to the model \({\bf{m}}\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = mg - b}}{{\bf{v}}^{\bf{r}}}{\bf{,v(0) = }}{{\bf{v}}_{\bf{o}}}\) (14).To study the effect of changing the parameter r in (14), take \({\bf{m = 1,}}\,\,{\bf{g = 9}}{\bf{.81,}}\,\,{\bf{b = 2}}\) and \({{\bf{v}}_{\bf{o}}}{\bf{ = 0}}\).Then use the improved Euler’s method subroutine with \({\bf{h = 0}}{\bf{.2}}\) to approximate the solution to (14) on the interval \(0 \le {\bf{t}} \le 5\) for \({\bf{r = 1}}{\bf{.0,}}\,\,{\bf{1}}{\bf{.5}}\) and 2.0. What is the relationship between these solutions and the constant solution\({\bf{v(t) = }}{\left( {\frac{{{\bf{9}}{\bf{.81}}}}{{\bf{2}}}} \right)^{\frac{{\bf{1}}}{{\bf{r}}}}}\)?

When \({\bf{r = 1}}{\bf{.0}}\), the constant solution is \({\bf{v(t) = 4}}{\bf{.905}}\).

When \({\bf{r = }}\,\,{\bf{1}}{\bf{.5}}\), the constant solution is \({\bf{v(t) = }}2.887\).

When \({\bf{r = }}\,\,{\bf{2}}{\bf{.0}}\), the constant solution is \({\bf{v(t) = }}2.215\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Important hint

For the solution apply Euler’s formula.

Step 2: Find the value of temperature when \({\bf{r = 1}}{\bf{.0}}\).

The given equation is

\(\begin{array}{l}{\bf{v(0) = 0}}\\\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = 9}}{\bf{.81 - 2}}{{\bf{v}}^{\bf{r}}}\end{array}\)

Now apply improved Euler’s method subroutine with \({\bf{h = 0}}{\bf{.2}}\) and \({\bf{N = 25}}\).

\(\begin{array}{c}{\bf{f(t,T) = 9}}{\bf{.81 - 2v}}\\{\bf{G = f(t + h,v + hF)}}\\{\bf{ = 9}}{\bf{.81 - 2(v + 0}}{\bf{.2(9}}{\bf{.81 - 2v))}}\end{array}\)

Apply initial conditions \({\bf{t = }}{{\bf{t}}_{\bf{o}}}{\bf{ = 0,v = }}{{\bf{v}}_{\bf{o}}}{\bf{ = 0}}\).

\(\begin{array}{c}{\bf{F(0,0) = 0}}{\bf{.2}}\\{\bf{G(0,0) = 5}}{\bf{.886}}\\{\bf{t = }}{{\bf{t}}_{\bf{0}}}{\bf{ + h = 0}}{\bf{.2}}\\{\bf{T = }}{{\bf{T}}_{\bf{o}}}{\bf{ + h}}\frac{{{\bf{F + G}}}}{{\bf{2}}}{\bf{ = 1}}{\bf{.570}}\end{array}\)

Therefore, at \({\bf{x = 0}}{\bf{.2}}\), the solutions are approximately \({\bf{v = 1}}{\bf{.570}}\).

The other values are

t

v

t

v

t

v

0.4

2.637

2

4.801

3.6

4.900

0.6

3.363

2.2

4.834

3.8

4.902

0.8

3.856

2.4

4.857

4

4.903

1

4.192

2.6

4.872

4.2

4.904

1.2

4.420

2.8

4.883

4.4

4.904

1.4

4.575

3

4.890

4.6

4.904

1.6

4.681

3.2

4.895

4.8

4.905

1.8

4.753

3.4

4.898

5

4.905

Step 3: Determine the value of temperature when\({\bf{r = 1}}{\bf{.5}}\).

The given equation is

\(\begin{array}{c}{\bf{v(0) = 0}}\\\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = 9}}{\bf{.81 - 2}}{{\bf{v}}^{\bf{r}}}\end{array}\)

Now apply improved Euler’s method subroutine with \({\bf{h = 0}}{\bf{.2}}\) and N=25.

\(\begin{array}{c}{\bf{f(t,T) = 9}}{\bf{.81 - 2}}{{\bf{v}}^{{\bf{1}}{\bf{.5}}}}\\{\bf{G = f(t + h,v + hF)}}\\{\bf{ = 9}}{\bf{.81 - 2(v + 0}}{\bf{.2(9}}{\bf{.81 - 2}}{{\bf{v}}^{{\bf{1}}{\bf{.5}}}}{\bf{)}}{{\bf{)}}^{{\bf{1}}{\bf{.5}}}}\end{array}\)

Apply initial conditions \({\bf{t = }}{{\bf{t}}_{\bf{o}}}{\bf{ = 0,v = }}{{\bf{v}}_{\bf{o}}}{\bf{ = 0}}\).

\(\begin{array}{c}{\bf{F(0,0) = 9}}{\bf{.81}}\\{\bf{G(0,0) = 4}}{\bf{.3136}}\\{\bf{t = }}{{\bf{t}}_{\bf{0}}}{\bf{ + h = 0}}{\bf{.2}}\\{\bf{T = }}{{\bf{T}}_{\bf{o}}}{\bf{ + h}}\frac{{{\bf{F + G}}}}{{\bf{2}}}\\{\bf{ = 1}}{\bf{.412}}\end{array}\)

Thus, at \({\bf{x = 0}}{\bf{.2}}\), the solutions are approximately \({\bf{v = 1}}.412\).

The other values are

t

v

t

v

t

v

0.4

2.150

2

2.884

3.6

2.887

0.6

2.519

2.2

2.885

3.8

2.887

0.8

2.703

2.4

2.886

4

2.887

1

2.795

2.6

2.887

4.2

2.887

1.2

2.841

2.8

2.887

4.4

2.887

1.4

2.864

3

2.887

4.6

2.887

1.6

2.875

3.2

2.887

4.8

2.887

1.8

2.881

3.4

2.887

5

2.887

Step 4: Evaluate the value of temperature when \({\bf{r = }}2.0\).

The given equation is

\(\begin{array}{c}{\bf{v(0) = 0}}\\\frac{{{\bf{dv}}}}{{{\bf{dt}}}}{\bf{ = 9}}{\bf{.81 - 2}}{{\bf{v}}^{\bf{r}}}\end{array}\)

Now apply improved Euler’s method subroutine with \({\bf{h = 0}}{\bf{.2}}\) and \({\bf{N = 25}}\).

\(\begin{array}{c}{\bf{f(t,T) = 9}}{\bf{.81 - 2}}{{\bf{v}}^{\bf{2}}}\\{\bf{G = f(t + h,v + hF)}}\\{\bf{ = 9}}{\bf{.81 - 2(v + 0}}{\bf{.2(9}}{\bf{.81 - 2}}{{\bf{v}}^{\bf{2}}}{\bf{)}}{{\bf{)}}^{\bf{2}}}\end{array}\)

Apply initial conditions\({\bf{t = }}{{\bf{t}}_{\bf{o}}}{\bf{ = 0,v = }}{{\bf{v}}_{\bf{o}}}{\bf{ = 0}}\).

\(\begin{array}{c}{\bf{F(0,0) = 9}}{\bf{.81}}\\{\bf{G(0,0) = 2}}{\bf{.111}}\\{\bf{t = }}{{\bf{t}}_{\bf{0}}}{\bf{ + h = 0}}{\bf{.2}}\\{\bf{T = }}{{\bf{T}}_{\bf{o}}}{\bf{ + h}}\frac{{{\bf{F + G}}}}{{\bf{2}}}{\bf{ = 1}}{\bf{.192}}\end{array}\)

Hence, at \({\bf{x = 2}}{\bf{.0}}\), the solutions are approximately \({\bf{v = 1}}.192\).

The other value is

t

v

t

v

t

v

0.4

1.533

2

2.218

3.6

2.201

0.6

1.719

2.2

2.146

3.8

2.204

0.8

1.841

2.4

2.160

4

2.206

1

1.927

2.6

2.171

4.2

2.208

1.2

1.991

2.8

2.180

4.4

2.209

1.4

2.039

3

2.187

4.6

2.210

1.6

2.077

3.2

2.193

4.8

2.211

1.8

2.105

3.4

2.197

5

2.211

Therefore, when \({\bf{r = 1}}{\bf{.0}}\), the constant solution is \({\bf{v(t) = 4}}{\bf{.905}}\).

When \({\bf{r = }}\,{\bf{1}}{\bf{.5}}\), the constant solution is \({\bf{v(t) = }}2.887\).

When \({\bf{r = }}\,{\bf{2}}{\bf{.0}}\), the constant solution is \({\bf{v(t) = }}2.215\).

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