6.1 slope fields and euler’s method quick review
TRANSCRIPT
6.1
Slope Fields and Euler’s Method
Quick Review
3
2
3 3
Determine whether or not the function satisfiesthe differential equation.
1. 3
2. csc 5 cot 5 3
13.
x
x
y
dy y y edxdy x y x xdxdy yydx
4
Find the constant C.4. 4 2 and 5 when 15. tan and 2 when 0x
y x x C y xy e x C y x
Quick Review Solutions
3
2
Determine whether or not the function satisfiesthe differential equation.
1. 3
2. csc 5 cot 5 3
Yes
Yes
13.
x
y
dy y y edxdy x y x xdxdy
ydx
3 3 No xy
4
Find the constant C.4. 4 2 and 5 when 15. tan and 2 when 0x
y x x C y xy e x C y x
C = 3C = – 3
What you’ll learn about Differential Equations Slope Fields Euler’s Method
Essential QuestionHow do we use differential equations andvisually demonstrate them using slopefields?
Differential EquationAn 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
.cos3satisfy that functions theall Find 1. 2 xxdx
dyy
y 3x xsin C
First-order Differential EquationIf the general solution to a first-order differential
equation is continuous, the only additional
information needed to find a unique solution 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 an Initial Value Problem
2. Find the particular solution to the equation
and y = 2 when x = 0.
xdx
dysin3
y 3 xcos C Cx cos3
2 C 0cos32 C 3
5C5cos3 xy
Example Solving an Initial Value Problem
3. Solve the initial value problem explicitly
y = 3 when x = 1.
1231
42
xxdx
dy
y 1x 3 x Cx 12 Cxxx
1211
3
3 C 1121
1
1
13
3 C14
11C 111211
3 x
xxy
2 x 123 4 x
Example Solving an Initial Value Problem
4. Find the particular solution to the equation
whose graph passes through the point (1, ½ ).
xedx
dy x 3 2
y xe2
2
1 2
2
3x C
2
1 12
2
1e C 21
2
3
2
1Ce
2
3
2
1 2
2
2
12 eC 222
2
12
2
3
2
1exey x
Example Using the Fundamental Theorem to Solve an Initial Value Problem
5. Find the solution to the differential equation
for which f (3) = 5.
2cos xxf
xf x
3 dtt cos 2 C
5 3
3
Cdtt cos 2
05C
x
dttxf
3
2 5 cos
Slope FieldThe differential equation gives the slope at any
point (x, y). This information can be used to
draw a small piece of the linearization at that
point, which approximates the solution curve
that passes through that point. Repeating that
process at many points yields an approximation
called a slope field.
Example Constructing a Slope Field6. Construct a slope field for the differential equation .cos x
dx
dy
At any point (0, y), the slope, dy/dx = cos 0 = .1
At any point (, y) or (, y), the slope, dy/dx = .1
The slope of odd multiples of /2 will be .0At any point (2, y) or (, y), the slope, dy/dx = .1
If you connect these, the graph of the f (x) should appear.
Cxxf sin
Euler’s Method for Graphing a Solution to an Initial Value Problem
1. Begin at the point ( x, y ) specified by the initial condition.
2. Use the differential equation to find the slope dy /dx at the point.
3. Increase x by x. Increase y by y, where y = (dy/dx) x. This defines a new point ( x + x, y + y) that lies along the linearization.
4. Using this new point, return to step 2. Repeating the process constructs the graph to the right of the initial point.
5. To construct the graph moving to the left from the initial point, repeat the process using negative values for x.
Example Applying Euler’s Method
1dy
xdx
,x y xdy
y xdx
,x x y y
Euler's Method leads us to the approximation (1.5) 3.1.f
7. Use Euler’s method with increments of x = 0.1 to approximate the value of y when x = 1.5 given
.1 when 2 and 1 xyxdx
dy
2 1.0 2 ,1 2.0 2.2 ,1.11.2 1.0 2.2 ,1.1 21.0 41.2 ,2.12.2 1.0 41.2 ,2.1 22.0 63.2 ,3.13.2 1.0 63.2 ,3.1 23.0 86.2 ,4.14.2 1.0 86.2 ,4.1 24.0 1.3 ,5.1
Pg. 327, 6.1 #1-49 odd