8.1.2 Euler's Method and Backwards Euler 

Given with use Euler's Method to approximate  

with and  

Then do the same using Backwards Euler's 

It's always a good idea to begin your page with a "restart" command -- so when re-running 

commands, you can start with a blank slate. 

>

restart;

 

>	f := (t,y) -> 5*t - 3*sqrt(y);

 

(1)

 

Let's approximate using a step size of . This means we need iterations. 

>	N := 2; h := 0.05;

 

 

	(2)

 

>	# Euler's Method t[0] := 0: y[0] := 2: for n from 1 to N do   t[n] := t[n-1] + h:   y[n] := y[n-1] + f(t[n-1],y[n-1])*h: end do:

 

So is approximately... 

>	yApprox := evalf(y[N]);

 

(3)

 

Now let's approximate and using a step size of . For we 

need iterations (is the value after 4 iterations and is the value after 6 iterations).
 

>	N := 8; h := 0.05;

 

 

	(4)

 

>	# Euler's Method t[0] := 0: y[0] := 2: for n from 1 to N do   t[n] := t[n-1] + h:   y[n] := y[n-1] + f(t[n-1],y[n-1])*h: end do:

 

So is approximately... 

>	yApprox := evalf(y[4]);

 

(5)

 

is approximately... 

>	yApprox := evalf(y[6]);

 

(6)

 

is approximately... 

>	yApprox := evalf(y[8]);

 

(7)

 

Next, let's approximate and using a step size of . For we 

need iterations (is the value after 4 iterations, is the value after 8
iterations and is the value after iterations). 

NOTE: Because trying to do all of the intermediate calculations exactly is too hard for my computer 

           I will introduce an "evalf" command in the next to last line of Euler's Method. This will approximate 

           the numbers used in the intermediate steps and greatly speed up computation.
 

>	N := 16; h := 0.025;

 

 

	(8)

 

>	# Euler's Method t[0] := 0: y[0] := 2: for n from 1 to N do   t[n] := t[n-1] + h:   y[n] := evalf(y[n-1] + f(t[n-1],y[n-1])*h): end do:

 

So is approximately... 

>	yApprox := evalf(y[4]);

 

(9)

 

is approximately... 

>	yApprox := evalf(y[8]);

 

(10)

 

is approximately... 

>	yApprox := evalf(y[12]);

 

(11)

 

is approximately... 

>	yApprox := evalf(y[16]);

 

(12)

 

Now for Backwards Euler's Method. 

Let's approximate and using Backwards Euler's and a step size of .  

For we need iterations (is the value after 2 iterations, is the value after 4
iterations, and is the value after iterations). 

NOTE: Because trying to do all of the intermediate calculations exactly is too hard for my computer 

           I will introduce an "evalf" command in the next to last line of Backwards Euler's. This will  

           approximate the numbers used in the intermediate steps and greatly speed up computation.
 

>	N := 8; h := 0.05;

 

 

	(13)

 

>	# Backwards Euler's Method t[0] := 0: y[0] := 2: for n from 1 to N do   t[n] := t[n-1] + h:   y[n] := evalf(solve(X = y[n-1] + f(t[n],X)*h, X)): end do:

 

So is approximately... 

>	yApprox := evalf(y[2]);

 

(14)

 

is approximately... 

>	yApprox := evalf(y[4]);

 

(15)

 

is approximately... 

>	yApprox := evalf(y[6]);

 

(16)

 

is approximately... 

>	yApprox := evalf(y[8]);

 

(17)

 

Finally, let's approximate and using Backwards Euler's and a step size of
. For we need iterations (is the value after 4 iterations, 

is the value after 8 iterations, and is the value after iterations). 

NOTE: Because trying to do all of the intermediate calculations exactly is too hard for my computer 

           I will introduce an "evalf" command in the next to last line of Backwards Euler's. This will  

           approximate the numbers used in the intermediate steps and greatly speed up computation.
 

>	N := 16; h := 0.025;

 

 

	(18)

 

>	# Backwards Euler's Method t[0] := 0: y[0] := 2: for n from 1 to N do   t[n] := t[n-1] + h:   y[n] := evalf(solve(X = y[n-1] + f(t[n],X)*h, X)): end do:

 

So is approximately... 

>	yApprox := evalf(y[4]);

 

(19)

 

is approximately... 

>	yApprox := evalf(y[8]);

 

(20)

 

is approximately... 

>	yApprox := evalf(y[12]);

 

(21)

 

is approximately... 

>	yApprox := evalf(y[16]);

 

(22)