BC 2-3
Euler, etc.
Name:
Problem 1
Back to BC 1-2 for a few moments…
Let y' = x + 1, y(0) = 0, for 0  x  6, step size = 2.
Sketch the piecewise-constant approximation for y' and then graph the resulting
piecewise-linear approximation for y.
y'
y
x
Remember, new y = old y + y, where y = rate  x
x
Problem 2
Again, let y' = x + 1, but this time, let y(–2) = 0, for –2  x  3.
Sketch the resulting piecewise-linear graph of y (only), using step size = 1.
Fill in the chart below to confirm your work above. This has been started for you.
( x, y )
( –2, 0)
rate = y' = x + 1
–1
rate  x = y
–1  1 = y
old y + y = new y
0 + (–1) = –1
(–1, –1)
Euler p.1
Fall 12
Problem 3
Let y' = y, with y(0) = 1 and x = 1/2. The process here is the same, with the exception
that we are not able to draw a piecewise-constant approximation for y'. (Why not?)
Complete a few lines of the following chart. This has been started for you.
( x, y )
(0, 1 )
rate = y'
1
rate  x = y
1 1/2 = 1/2
old y + y = new y
1 + 1/2 = 3/2
( 1/2, 3/2)
Sketch the slope field for y' = y on your calculator. Do your few points above seem to
agree with the slope field?
Now do Euler on your calculator. Start with the point (0, 1). Does your graph agree?
Repeat the graph with different initial values.
Euler p.2
Fall 12
Problem 4
Let y' = ƒ(x, y) = x + y. The slope field is shown below.
3
Draw approximate solution curves for
each of the following starting points.
How does each one differ from the
others?
2
1
(–3, 2.5)

4
2
2
4
1
(–3, 2)
2
(–3, 1.5)
3
Use Euler on your calculator to see if your solution graphs are correct (or at least close!).
Create a few steps of the chart for initial point (–3, 2.5) and step size x = 0.2
( x,
y )
rate = y'
rate  x = y
old y + y = new y
Check these values with Euler on your calculator. (Use Trace on the 89.) Do these
values agree?
Euler p.3
Fall 12