Math 3C Euler’s Method Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.

Math 3C

Euler’s Method

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Euler’s Method will find an approximate solution to an initial value problem.

Let’s work through a simple example:

Suppose we want to solve the following initial value problem:

1)0(y

y3.0y

a) First find an estimate for y(2) using Euler’s Method

with a step size of 1

b) Next get a better estimate by using a step size of 0.5.

c) Finally, solve the problem exactly and compare to the estimates from parts a) and b)






1)0(y

y3.0y

a) First find an estimate for y(2) using Euler’s Method

with a step size of 1



Euler’s method will create a piecewise linear approximation to y(t) by using the given differential equation to calculate the slope of the solution curve at each step, then following that slope to calculate the approximations.






a) First find an estimate for y(2) using Euler’s Method with a step size of 1



The first step is to use the given equation and initial value to find the initial slope (at t=0):

3.0)1(3.0)0(y3.0)0(y

1)0(y

y3.0y










Next we use this slope to find the approximation for y(1):

3.1)1()3.0(1)1(y~t)0(y)0(y)1(y~

1)0(y

y3.0y

3.0)1(3.0)0(y3.0)0(y











Now we repeat this process, calculating the slope at t=1:

39.0)3.1()3.0()1(y~)3.0()1(y~

1)0(y

y3.0y

3.0)1(3.0)0(y3.0)0(y

3.1)1()3.0(1)1(y~t)0(y)0(y)1(y~











Now we repeat this process, calculating the slope at t=1:

We have one last step – calculate the approximation for y(2):

69.1)1()39.0(3.1)2(y~t)1(y~)1(y~)2(y~

Here is our answer to part a)

1)0(y

y3.0y

3.0)1(3.0)0(y3.0)0(y

3.1)1()3.0(1)1(y~t)0(y)0(y)1(y~

39.0)3.1()3.0()1(y~)3.0()1(y~



It is helpful to see a graph of our results.

y(step size 1)

00.20.40.60.8

11.21.41.61.8

2

0 0.5 1 1.5 2

Notice that we used two steps, and we have two straight line segments

(whose slopes we calculated from the given D.E.)

Slope=0.3

Slope=0.39



Now we can run through the procedure for part b). We use 4 steps this time, each of step size 0.5.

This time let’s make a table to organize our calculations:

t y y’

0.0 1.0

0.5

1.0

1.5

2.0





t y y’

0.0 1.0

0.5

1.0

1.5

2.0

Calculate y’(0) from the original D.E.

y3.0y





t y y’

0.0 1.0 0.3

0.5

1.0

1.5

2.0

Calculate y’(0) from the original D.E.

y3.0y





t y y’

0.0 1.0 0.3

0.5

1.0

1.5

2.0

Find y(0.5) from the formula

t)0(y)0(y)5.0(y~





t y y’

0.0 1.0 0.3

0.5 1.15

1.0

1.5

2.0


15.1)5.0()3.0(1)5.0(y~t)0(y)0(y)5.0(y~





t y y’

0.0 1.0 0.3

0.5 1.15

1.0

1.5

2.0

Calculate y’(0.5) from the original D.E.

y3.0y





t y y’

0.0 1.0 0.3

0.5 1.15 0.345

1.0

1.5

2.0

Calculate y’(0.5) from the original D.E.

345.0)15.1()3.0(y





t y y’

0.0 1.0 0.3

0.5 1.15 0.345

1.0

1.5

2.0


t)5.0(y~)5.0(y~)0.1(y~





t y y’

0.0 1.0 0.3

0.5 1.15 0.345

1.0 1.3225

1.5

2.0


3225.1)0.1(y~)5.0()345.0(15.1)0.1(y~

t)5.0(y~)5.0(y~)0.1(y~





t y y’

0.0 1.0 0.3

0.5 1.15 0.345

1.0 1.3225

1.5

2.0

Fill in the rest of the table in the same manner, calculating a slope, then using it to get the next value for y…





t y y’

0.0 1.0 0.3

0.5 1.15 0.345

1.0 1.3225 0.39675

1.5 1.520875

2.0


I kept all the digits, but you can

probably round off a little bit





t y y’

0.0 1.0 0.3

0.5 1.15 0.345

1.0 1.3225 0.39675

1.5 1.520875 0.4562625

2.0 1.74900625


I kept all the digits, but you can

probably round off a little bit

Here is our answer to part b)



Here is the graph of our results from parts a) and b).

The new calculation should be more accurate because we used more steps that were closer together. In the next step we will find the exact answer.

2, 1.69

2, 1.749

00.20.40.60.8

1

1.21.41.61.8

2

0 0.5 1 1.5 2

y(step size 1) y(step size 0.5)



1)0(y

y3.0y

We want the value for y(2).

First we will separate, then integrate to find the general solution.

dt3.0y3.0y

dydtdy

We can actually solve this problem, so let’s do it using separation of variables.

(You should be able to find the general solution to this one in your head in <5 seconds, btw).



1)0(y

y3.0y



dt3.0y3.0y

dydtdy

t3.0CeyCt3.0yln





1)0(y

y3.0y



dt3.0y3.0y

dydtdy

t3.0CeyCt3.0yln

Next put in the given initial value to find C:





1)0(y

y3.0y



dt3.0y3.0y

dydtdy

t3.0CeyCt3.0yln


1CCe1Ce)0(y 0)0(3.0





1)0(y

y3.0y



dt3.0y3.0y

dydtdy

t3.0CeyCt3.0yln


1CCe1Ce)0(y 0)0(3.0

So our solution is t3.0e)t(y





1)0(y

y3.0y





dt3.0y3.0y

dydtdy

t3.0CeyCt3.0yln


1CCe1Ce)0(y 0)0(3.0

So our solution is t3.0e)t(y

Last step – plug in t=2

82212.1e)2(y 6.0



Here is the graph comparing our results.

Notice that the approximation with the smaller step size is closer to the exact value (but it took more work to get there).

We can calculate the % error in our estimates as a check on their accuracy.

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 1 2 3

y(step size 1)

y(step size 0.5)

y=e (̂0.3t)

y(2) % error

step size =1

1.690 7.2%

step size =0.5

1.749 4.0%

exact 1.822 0.0%



Math 3C Euler’s Method Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.

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