(a) Use Euler's method with each of the following step sizes...

(a) Use Euler's method with each of the following step sizes to estimate the value of $ y(0.4), $ where $ y $ is the solution of the initial-value problem $ y' = y, y(0) = 1. $ (i) $ h = 0.4 $ (ii) $ h = 0.2 $ (iii) $ h = 0.1 $ (b) We know that the exact solution of the initial-value problem in part (a) is $ y = e^x. $ Draw, as accurately as you can, the graph of $ y = e^x, 0 \le x \le 0.4, $ together with the Euler approximations using the step sizes in part (a). (Your sketches should resemble Figure 12, 13, and 14.) Use your sketches to decide whether your estimates in part (a) are underestimates or overestimates. (c) The error in Euler's method is the difference between the exact value and the approximate value. Find the errors made in part (a) in using Euler's method to estimate the true value of $ y(0.4), $ namely $ e^{0.4}. $ What happens to the errors each time the steps size is halved?

(a) Use Euler's method with each of the following step sizes to estimate the value of $ y(0.4), $ where $ y $ is the solution of the initial-value problem $ y' = y, y(0) = 1. $
(i) $ h = 0.4 $ (ii) $ h = 0.2 $ (iii) $ h = 0.1 $
(b) We know that the exact solution of the initial-value problem in part (a) is $ y = e^x. $ Draw, as accurately as you can, the graph of $ y = e^x, 0 \le x \le 0.4, $ together with the Euler approximations using the step sizes in part (a). (Your sketches should resemble Figure 12, 13, and 14.) Use your sketches to decide whether your estimates in part (a) are underestimates or overestimates.
(c) The error in Euler's method is the difference between the exact value and the approximate value. Find the errors made in part (a) in using Euler's method to estimate the true value of $ y(0.4), $ namely $ e^{0.4}. $ What happens to the errors each time the steps size is halved?

Transcript

00:02 In part a, we are given a differential equation with an initial value, and we are asked to use spoilers method to approximate the value of a function y at 0 .4.

00:20 Differential equation is y -prime equals y.

00:24 Initial value is y of 0 equals 1.

00:30 In part 1, or part a, we are asked to approximate y of 0 .4.

00:38 Given a step size h of 0 .4.

00:51 So using the layer method, you have the y of 1 is equal to y of 0, which is the same thing.

01:05 Y 0 is the same as y of 0 or 1, plus a step size h times our function y, the value we did it the point 0 ,0 ,1, which is simply 1.

01:34 This is equal to 1 plus 0 .4 or 1 .4.

01:45 Y2 is equal to y1, which you have is 1 .4, plus our step size, 0 .4 times, okay, so you don't need to do a lot of 2.

02:08 Because y of 1 is already an approximation for y of 0 .4 since we move from 0 to 0 .4 this is approximately equal to y of 1 which is equal to 1 .4 so answer for part 1 in part 2 you're asked to approximate this value using a step size h of line 0 .2 and we see that again, y1 is 1 to be equal to y0 for 1 plus the set size .2 times our function y, divided it at the point 0 .01, which is just 1.

03:14 This is equal to 1 .2.

03:18 In order to get from x equals 0 to x to 0, we need to add another set.

03:25 So we have y of 2 is equal to y of 1 which is 1 .2 plus the step size .2 times our function y evaluated at the point.

03:45 1 .2, 1 .2.

03:49 And this can be 1 .2.

04:00 This is equal to 1 .2 plus .2 times 1 .2 times 1 .2.

04:13 Is 1 .44.

04:17 So we see that with this step size, y of 0 .4, is approximately equal to y of 2, which is equal to 1 .44.

04:39 And as i answer to part 2, in part 3, you're asking you to set size of 0 .1.

04:50 There will be 4 steps.

04:51 So y1, point equal to y0 to 1 plus our step size, 0 .1 times the function y is how to be it to 0 .1, so 01, which is 1.

05:17 So this is 1 plus 0 .1 for 1 .1.

05:23 Second step, y2, when the value is going to be y1, 2 plus our step 11, plus our step size, 0 .1 .1.

05:33 Times the value of function y at the point 0 .1 .1 is simply 1 .1.

05:44 This is equal to 1 .1 plus .1 times 1 .1 is equal to 1 .21.

06:04 Y3 is equal to y2, for 1 .21, plus that step size times.

06:10 The value of the function y at the point point two, 1 .21, which is going to be 1 .21, and this is equal to 1 .331.

06:37 And finally, for our last step, we have the value y4 is equal to y3, or 1 .331, plus 0 .1 times.

06:53 The value of the function y at the point.

06:58 0 .3, 1 .331, which is 1 .331, which is equal to 1 .4641.

07:18 And so we have that y.

07:24 0 .4 is approximately equal to y4, which is equal to 1 .4641.

07:37 This is our answer to part three.

07:46 In part b, we are given the exact solution of our initial value problem.

08:00 We are asked to draw the graphs of the solution using our oiler approximations.

08:23 And then we're asked to evaluate whether or not our approximations are under or over estimates of the true value.

08:36 So we're giving that the solution to our mutual value problem is y equals b to the x we're asked to draw this from 0 to 0 on the x axis.

09:13 Now we know this is all going to be in the first quadrant so i'm going to draw the first quadrant and we're going to make this as detailed as i can because otherwise this isn't going to be very useful so i'm going to make the largest value on the x -axis.

09:30 One to be .4, so we also have .2 here, point 3 here, and 0 .1 here.

09:47 This is 0.

09:51 In the y direction, you have that, i just draw this table of values here...

Please give Ace some feedback