4.2 Direction fields and numerical methods (Page 5/14)

Calculus volume 2 Page 5 / 14

Typically $h$ is a small value, say $0.1$ or $0.05 .$ The smaller the value of $h,$ the more calculations are needed. The higher the value of $h,$ the fewer calculations are needed. However, the tradeoff results in a lower degree of accuracy for larger step size, as illustrated in [link] .

Two graphs of the same parabola, y = x ^ 2 – 3 x + 3. The first shows Euler’s method for the given initial-value problem with a step size of h = 0.05, and the second shows Euler’s method with a step size of h = 0.25. The first then has the points (0, 3), (.5, 1.5), (1, 0.5), (1.5, 0), (2, 0), (2.5, 0.5), and (3, 1.5) plotted with line segments connecting them. The second has the points (0, 3), (0.25, 2.25), (0.5, 1.625), (0.75, 1.125), (1, 0.75), (1.25, 0.5), (1.5, 0.375), (2, 0.5), (2.25, 0.75), (2.5, 1.125), (2.75, 1.625), and (3, 2.25) plotted with line segments connecting them. — Euler’s method for the initial-value problem $y^{'} = 2 x - 3, y (0) = 3$ with (a) a step size of $h = 0.5;$ and (b) a step size of $h = 0.25 .$

Using euler’s method

Consider the initial-value problem

y^{'} = 3 x^{2} - y^{2} + 1, y (0) = 2 .

Use Euler’s method with a step size of $0.1$ to generate a table of values for the solution for values of $x$ between $0$ and $1 .$

We are given $h = 0.1$ and $f (x, y) = 3 x^{2} - y^{2} + 1 .$ Furthermore, the initial condition $y (0) = 2$ gives $x_{0} = 0$ and $y_{0} = 2 .$ Using [link] with $n = 0,$ we can generate [link] .

Using euler’s method to approximate solutions to a differential equation
$n$	$x_{n}$	$y_{n} = y_{n - 1} + h f (x_{n - 1}, y_{n - 1})$
$0$	$0$	$2$
$1$	$0.1$	$y_{1} = y_{0} + h f (x_{0}, y_{0}) = 1.7$
$2$	$0.2$	$y_{2} = y_{1} + h f (x_{1}, y_{1}) = 1.514$
$3$	$0.3$	$y_{3} = y_{2} + h f (x_{2}, y_{2}) = 1.3968$
$4$	$0.4$	$y_{4} = y_{3} + h f (x_{3}, y_{3}) = 1.3287$
$5$	$0.5$	$y_{5} = y_{4} + h f (x_{4}, y_{4}) = 1.3001$
$6$	$0.6$	$y_{6} = y_{5} + h f (x_{5}, y_{5}) = 1.3061$
$7$	$0.7$	$y_{7} = y_{6} + h f (x_{6}, y_{6}) = 1.3435$
$8$	$0.8$	$y_{8} = y_{7} + h f (x_{7}, y_{7}) = 1.4100$
$9$	$0.9$	$y_{9} = y_{8} + h f (x_{8}, y_{8}) = 1.5032$
$10$	$1.0$	$y_{10} = y_{9} + h f (x_{9}, y_{9}) = 1.6202$

With ten calculations, we are able to approximate the values of the solution to the initial-value problem for values of $x$ between $0$ and $1 .$

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Go to this website for more information on Euler’s method.

Consider the initial-value problem

y^{'} = x^{3} + y^{2}, y (1) = −2 .

Using a step size of $0.1,$ generate a table with approximate values for the solution to the initial-value problem for values of $x$ between $1$ and $2 .$

$n$	$x_{n}$	$y_{n} = y_{n - 1} + h f (x_{n - 1}, y_{n - 1})$
$0$	$1$	$−2$
$1$	$1.1$	$y_{1} = y_{0} + h f (x_{0}, y_{0}) = −1.5$
$2$	$1.2$	$y_{2} = y_{1} + h f (x_{1}, y_{1}) = −1.1419$
$3$	$1.3$	$y_{3} = y_{2} + h f (x_{2}, y_{2}) = −0.8387$
$4$	$1.4$	$y_{4} = y_{3} + h f (x_{3}, y_{3}) = −0.5487$
$5$	$1.5$	$y_{5} = y_{4} + h f (x_{4}, y_{4}) = −0.2442$
$6$	$1.6$	$y_{6} = y_{5} + h f (x_{5}, y_{5}) = 0.0993$
$7$	$1.7$	$y_{7} = y_{6} + h f (x_{6}, y_{6}) = 0.5099$
$8$	$1.8$	$y_{8} = y_{7} + h f (x_{7}, y_{7}) = 1.0272$
$9$	$1.9$	$y_{9} = y_{8} + h f (x_{8}, y_{8}) = 1.7159$
$10$	$2$	$y_{10} = y_{9} + h f (x_{9}, y_{9}) = 2.6962$

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Visit this website for a practical application of the material in this section.

Key concepts

A direction field is a mathematical object used to graphically represent solutions to a first-order differential equation.
Euler’s Method is a numerical technique that can be used to approximate solutions to a differential equation.

Key equations

Euler’s Method
$\begin{matrix} x_{n} = x_{0} + n h \\ y_{n} = y_{n - 1} + h f (x_{n - 1}, y_{n - 1}), where h is the step size \end{matrix}$

For the following problems, use the direction field below from the differential equation $y' = −2 y .$ Sketch the graph of the solution for the given initial conditions.

$y (0) = 1$

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$y (0) = 0$

A graph of the given direction field with a flat line drawn on the axis. The arrows point up for y < 0 and down for y > 0. The closer they are to the x axis, the more horizontal the arrows are, and the further away they are, the more vertical they become.

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$y (0) = −1$

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Are there any equilibria? What are their stabilities?

$y = 0$ is a stable equilibrium

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For the following problems, use the direction field below from the differential equation $y' = y^{2} - 2 y .$ Sketch the graph of the solution for the given initial conditions.
A direction field with horizontal arrows at y = 0 and y = 2. The arrows point up for y > 2 and for y < 0. The arrows point down for 0 < y < 2. The closer the arrows are to these lines, the more horizontal they are, and the further away, the more vertical the arrows are.

$y (0) = 3$

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$y (0) = 1$

A direction field with horizontal arrows at y = 0 and y = 2. The arrows point up for y > 2 and for y < 0. The arrows point down for 0 < y < 2. The closer the arrows are to these lines, the more horizontal they are, and the further away, the more vertical the arrows are. A solution is sketched that follows y = 2 in quadrant two, goes through (0, 1), and then follows the x axis.

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$y (0) = −1$

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Are there any equilibria? What are their stabilities?

$y = 0$ is a stable equilibrium and $y = 2$ is unstable

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Draw the direction field for the following differential equations, then solve the differential equation. Draw your solution on top of the direction field. Does your solution follow along the arrows on your direction field?

$y' = t^{3}$

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$y' = e^{t}$

A direction field over the four quadrants. As t goes from 0 to infinity, the arrows become more and more vertical after being horizontal closer to x = 0.

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$\frac{d y}{d x} = x^{2} cos x$

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$\frac{d y}{d t} = t e^{t}$

A direction field over [-2, 2] in the x and y axes. The arrows point slightly down and to the right over [-2, 0] and gradually become vertical over [0, 2].

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$\frac{d x}{d t} = cosh (t)$

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Draw the directional field for the following differential equations. What can you say about the behavior of the solution? Are there equilibria? What stability do these equilibria have?

Questions & Answers

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Prevent foreign microbes to the host

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is the fundamental units of Life

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Binomial nomenclature

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Source: OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2

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