### Topics

## Introduction to Differential Equations - At A Glance:

Things to remember about Euler's Method:

- Euler's Method gives only approximate values unless the function happens to be a straight line, in which case Euler's Method gives exact values of the function.

- Euler's method does NOT produce a formula. It generates a
*numerical solution*; that is, approximate values of the function at certain points.

The **error** in an approximation is the difference between the approximation and the real value of the function.

We can only figure out the error if we know the real value of the function.

### Sample Problem

Our first examples of Euler's method used the function *y* that satisfied the IVP

By thinking backwards we can figure out that

*y* = *x*^{2}.

Now we can compute the real value of *y* at 1:

*y*(1) = (1)^{2} = 1.

This lets us figure out how bad our approximations were in some earlier examples.

Using a step size of 0.5 we estimated

*y*(1) ≅ 0.5.

This has an error of 0.5, since it's 0.5 off from the real answer *y*(1) = 1.

Using a step size of 0.25 we estimated

*y*(1) ≅ 0.75.

The error here is 0.25.

Using a step size of 0.2 we estimated

*y*(1) ≅ 0.8.

This has an error of 0.2.

#### Example 1

Let *y* = *f* (*x*) be a solution to the IVP Does Euler's method produce an over- or under-estimate for the value of *f* (2)? | |

We left out the number of steps to use with Euler's method, but that doesn't matter for the answer. The slope field for looks like this: Since the initial condition puts the solution under the line *y* = *x*, the solution is concave down: Therefore that Euler's method will produce an overestimate. | |

#### Example 2

Let *y* = *f* (*x*) be a solution to the IVP For which starting points will Euler's method produce underestimates and for which starting points will Euler's method produce overestimates? | |

The slope field looks like this: A non-zero solution to this d.e. must lie either entirely above or entirely below the *x*-axis. If a solution lies above, it is concave up. If a solution lies below, it is concave down. For any starting point of the form (*x*,*y*) where *y* > 0, Euler's method will produce an underestimate. For any starting point of the form (*x*,*y*) where *y* < 0, Euler's method will produce an overestimate. | |

#### Exercise 1

Let *y* = *f* (*x*) be a solution to the IVP

What is the error when Euler's method is used to estimate *f* (2) with a step size of 0.5? With a step size of 0.25?

Answer

Since we're asked about error, we'll need to know the exact value of *f* (2). Thinking backwards, *f* (*x*) = 2*x*^{2} + *C*. Since *f* (1) = 3,

2(1)^{2} + *C* = 3

so *C* = 1 and *f* (*x*) = 2*x*^{2} + 1. The exact value in question is

*f* (2) = 2(2)^{2} + 1 = 9.

Euler's method with step size 0.5 gets us this table:

The error is 1, since the exact value of *f* (2) is 9 but Euler's method approximated *f* (2) ≅ 8. Euler's method with step size .25 gets us this table:

With step size 0.25 Euler's method approximates *f* (2) ≅ 8.5. The error in this case is 0.5.

#### Exercise 2

If *f* is concave up around *x* = *a*, is the tangent line to *f* at a above or below the graph of *f*? What about if *f* is concave down?

Answer

If *f* is concave up, the tangent line to *f* at a is below the graph of *f*. This is true whether the slope of the tangent line is negative, zero, or positive.

The moral of the previous exercise is that when *f* is concave up, a tangent line approximation is an underestimate. Since Euler's method is a bunch of tangent line approximations stuck together, Euler's method will also provide an underestimate, regardless of how many steps are used

When *f* is concave down, a tangent line approximation is an overestimate. Since Euler's method is a sequence of tangent line approximations, Euler's method also provides an overestimate, regardless of how many steps are used.

Remember when we drew, on a slope field, solutions to a differential equation that passed through a particular point? If we can sketch a solution, we can tell whether it's concave up or concave down. That's all we need to tell whether Euler's method will overestimate or underestimate.

#### Exercise 3

Let *y* = *f* (*x*) be a solution to the IVP

Does Euler's method produce an over- or under-estimate for the value of *f* (2)?

Answer

If we draw the solution and slope field, we see that this solution is concave up:

This means Euler's method will produce an underestimate to the value *f* (2).

#### Exercise 4

Let *y* = *f* (*x*) be a solution to the IVP

Does Euler's method produce an over- or under-estimate for the value of *f* (3.5)?

Answer

We draw the slope field and the solution that passes through (3,1):

The solution is concave down, so Euler's method will produce an overestimate to the value *f* (3.5).

#### Exercise 5

Let *y* = *f* (*x*) be a solution to the IVP

Does Euler's method produce an over- or under-estimate for the value of *f* (1)?

Answer

Draw the solution to this differential equation that passes through (0,5):

This solution is concave up, so Euler's method will produce an underestimate to the value *f* (1).

#### Exercise 6

Let *y* = *f* (*x*) be a solution to the IVP

Does Euler's method produce an over- or under-estimate for the value of *f* (2)?

Answer

This is a trick question. The solution to that passes through the point (1,1) is the line *y* = *x*.

Since the tangent line to a line is that line, Euler's method in this case will produce the exact value of *f* (2).

#### Exercise 7

Let *y* = *f* (*x*) be a solution to the IVP

Does Euler's method produce an over- or under-estimate for the value of *f* (5.9)?

Answer

The solution that passes through (5,-1) is concave down.

Therefore Euler's method will produce an overestimate to *f* (5.9).