Drill: Find the differential of each equation.

transcript

• y = ex

• y = e4x

• y = ex^2

• y = sec x

• dy = ex dx• dy = 4e4x dx• dy = 2xex^2 dx• dy = tanxsecx dx

Slope Fields and Euler’s Method

Lesson 6.1

Objectives

• Students will be able to:– construct antiderivatives using the Fundamental Theorem

of Calculus.– solve initial value problems in the form– construct slope fields using technology and interpret slope

fields as visualizations of different equations.– use Euler’s Method for graphing a solution to an initial

value problem.

., 00 xfyxfdx

Differential Equation

• An equation involving a derivative is called a differential equation. The order of a differential equation is the order of the highest derivative involved in the equation.

Example Solving a Differential Equation

• Find all functions y that satisfy

• Solution: the anti-derivative of dy/dx:

• This is called a GENERAL solution.• We cannot find a UNIQUE solution unless

we are given further information.

52sec2 xxdxdy

Cxxxy 5tan 2

• If the general solution to the first order differential equation is continuous, the only information needed is the value of the function at a single point, called an INITIAL CONDITION.

• A differential equation with an initial condition is called an INITIAL VALUE PROBLEM.

• It has a unique solution, called the PARTICULAR SOLUTION to the differential equation.

Example Solving a Differential Equation

Find all the functions y that satisfy .1

Cxxy ln6

Example Solving an Initial Value Problem

Find the particular solution to the equation

whose graph passes through the point (0, 1).

,12 xdx

dxxy 12

Cxxy 3

1 3 xxy

Example Solving an Initial Value Problem

Find the particular solution to the equation

and y = 60 when x = 4.

,310 xdx

dxxy 21

Cxxy 23

4241060

4210 23

Example Handling Discontinuity

• Find the particular solution to the equation dy/dx = 2x – sec2x whose graph passes through the point (0, 3)

• The general solution is y = x2 – tanx + C• Applying the initial condition: 3 = 02 – tan0 + C; 3 = C• Therefore, the particular solution is y = x2 – tanx + 3• However, since because tanx is does have

discontinuities, you need to add the domain stipulation of –π/2 < x< π/2

Example Using the Fundamental Theorem

• Find the solution to the differential equation f’(x) = e-x^2 for which f(7) = 3

• Rather than determining the anti-derivative, you can let the solution in integral form:

• We know that if

• Then AND• f’(x) = e-x^2

3303)7(7

dtef t

t dtexf7

This is our solution!

Note: If you are ABLE to determine the anti-derivative, you should do so!

Drill: Find the constant C• y = 3x2 + 4x + C and y = 2

when x = 1

• y = 2sinx – 3cosx + C and y = 4 when x = 0

• 2 = 3(1)2 + 4(1) + C• 2=3+4+C• 2 = 7 + C• -5 = C

• 4 = 2sin(0) – 3cos(0) + C• 4 = 0 – 3 + C• 4 = -3 + C• 7 = C

Graphing a General Solution

• Graph the family of functions that solve the differential equation dy/dx = cos x

• Any function of the form y = sinx + C solves the differential equation.

• First graph y = sin x, and then repeat the graph by shifting vertically up and down.

Note: you can also use the calc by putting values of ‘C’ in to L1 and then graphing y1 = sinx + L1

Slope Fields

• a slope field (or direction field) is a graphical representation of the solutions of a first-order differential equation. It is achieved without solving the differential equation analytically. The representation may be used to qualitatively visualize solutions, or to numerically approximate them.

• Remember also, that the derivative of a function gives its slope.

• Also remember that we use the notation dy/dx to represent derivative; therefore, dy/dx = slope.

A Summary of Making Slope Fields

• Put the differential equation in the form dy/dx = g(x,y)

• Decide upon what rectangular region of the plane you want to make the picture

• Impose a grid on this region• Calculate the value of the slope, g(x,y), at each grid

point, (x,y)• Sketch a picture in which at each grid point there is a

short line segment having the corresponding slope

Example: Constructing a Slope Field

• Construct a slope field for the differential equation dy/dx = cos x

• We know that that slope at any point (0, y) will be cos (0) = 1, so we can start be drawing tiny segments with slope 1 at several points along the y-axis.

• Slope is also 1 at 2π, - 2π

• When would the slope by -1?– At π, - π

• When is the slope 0?– At π/2, - π/2– At 3π/2, - 3π/2

Solutionx: [-5π/2, 5π/2]; y: [-4, 4]

Example Matching Slope Fields with Differential Equations

Use slope analysis to match the differential equation with the given slope fields. Note: each block is .5 units.

Use slope analysis to match the differential equation with the given slope fields. Note: each block is .5 units

then,2 If

21 ydx

then,1

or 1 If

Constructing a Slope Field for a Nonexact Differential Equation

• Construct a slope field for the differential equation dy/dx = x + y and sketch a graph of the particular solution that passes through (2, 0).

• You can make tables in order to graph slopes: (some examples)

x + y = 0 x + y = -1 x + y = 1

• The particular solution can be found by drawing a smooth curve through the point (2, 0) that follows the slopes in the slope field.

x y-2 2-1 10 01 -12 -2

x y-2 1-1 00 -11 -22 -3

x y-2 3-1 20 11 02 -1

Solution http://www.math.rutgers.edu/~sontag/JODE/JOdeApplet.html

• Find a solution that satisfies y(1) = 2 when

• First, determine the anti-derivative.

Euler’s Method• Begin at the point (x, y) specified by the initial condition. This

point will be on the graph, as required.• Use the differential equation to find the slope dy/dx at the

point.• Increase x by a small amount of Δx. Increase y by a small

amount of Δy, where Δy = (dy/dx) Δx. This defines a new point (x + Δx, y + Δy) that lies along the linearization.

• Using this new point, return to step 2. Repeating the process constructs the graph to the right of the initial point.

• To construct the graph moving to the left from the initial point,r epeat the process using negative values for Δx.

Example Applying Euler’s Method

Let f be the function that satisfies the initial value

problem and f (0) = 1. Use Euler’s

Method and increments of Δx = 0.2 to approximate f (1).

2.0,2 xxydx

yx, 2xydx

dy x x

dyy yyxx ,

1,0 1 2.0 2.0 2.1,2.0

2.1,2.0 24.1 2.0 248.0 448.1,4.0

448.1,4.0 608.1 2.0 3216.0 7696.1,6.0

7696.1,6.0 1296.2 2.0 425592.0 19552.2,8.0

19552.2,8.0 83554.2 2.0 567104.0 762624.2,1

762624.21 f

Let f be the function that satisfies the initial value

problem and if y = 3 when x = 2, use

Euler’s Method with five equal steps to approximate y when x = 1.5.

Δx = (1.5 – 2)/5 = -.1

(Note, Δx is negative when you are going backwards.

1.,2 xyxdx

yx, yxdx

dy2 x x

dyy yyxx ,

3,2 1 1. 1. )9.2,9.1(

9.2,9.1 9. 1. 09. 81.2,8.1

81.2,8.1 79. 1. 079. 731.2,7.1

731.2,7.1 669. 1. 0669. 6641.2,6.1

6641.2,6.1 5359. 1. 05359. 61051.2,5.1

61.25.1 f

Homework

• Day #1:Page 327: 1-19: odd• Day #2: Page 327/8: 21-39: odd• Day #3: page 328: 41-48

Drill: Find the differential of each equation.

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