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Found in: Page 131

### 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

# 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 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 $${\bf{v(t) = }}2.215$$.