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Numerade Educator



Problem 25 Easy Difficulty

(a) Program a calculator or computer to use Euler's method to compute $ y(1), $ where $ y(x) $ is the solution of the initial-value problem
$ \frac {dy}{dx} + 3x^2y = 6x^2 y(0) = 3 $
(i) $ h = 1 $ (ii) $ h = 0.1 $
(iii) $ h = 0.01 $ (iv) $ h = 0.001 $
(b) Verify that $ y = 2 + e^{-x^3} $ is the exact solution of the differential equation.
(c) Find the errors in using Euler's method to compute $ y(1) $ with the step sizes in part (a). What happens to the error when the step size is divided by 10?


(i) $y(1)=3$
(ii) $y(1) \approx 2.3928$
(iii) $y(1) \approx 2.3701$
(iv)$y(1) \approx 2.3681$
b) $y(0)=2+e^{-0}=2+1=3$
c) $\ln (\text { ii })-(\text { iv }),$ it seems that when the step size is divided by $10,$ the error estimate is also divided by 10 (approximately).


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Video Transcript

Okay. And this problem we're using Oilers method to solve a differential equation. And that differential equation is, do I dx plus three X squared y equals six X squared. With the initial condition y of zero equals three. So the first thing we wanna do is we want to have the derivative by itself on the left hand side. So we'll rewrite this as dy dx equals six X squared -3 X Squared Why? And we'll take this as our F. Of X. Why? Okay then we'll use the formula and I'll do the formula for the first example because it's just one step. Why then equals Live N -1 plus each F of X & -1. Why? & -1. So from the initial condition we have zero equals zero. And wise. So to find the value after one step Which would be the value at x equals one. We can use the formula. Mhm. Like this. So if you plug in X equals zero and y equals three into F of X. Y. You'll get zero. This six X squared goes to zero and three X squared y goes to zero. So we get why one equals three plus one times zero which equals three. So for the first part we have why have one equals three. Okay, the next parts have more steps. So I wrote a computer program in Mathematica in order to solve those parts. So if you take a look at the form that I have here, we have the initial condition we have our step size and then we have our differential equation. Oilers method is right below it. And the output is the answer that will get. So when H equals .1 We have y equals 2.39 to eat. When H. equals .01. Yeah Y equals 2.3701. And for the last part when h equals .001 we have Y equals 2.368 one. Okay. Yeah part B asked us to verify that. Mm Yeah Y equals two plus E. To the negative X cube is an exact solution to the differential equation. Dy dx equals six X squared minus three X squared. Why? With the initial condition wives zero equals three. Okay, so the left hand side is the derivative. Yeah. Of why? Which is two plus E. To the negative X cubed. Using the chain rule we get negative three X squared E. To negative X. Cute. For the right hand side we plug in. Why? To get six X squared minus three X squared. Why? Which is two plus eaten negative x cubed. And when you simplify it you again get negative three X squared E. To the negative X cubed. So these two are equal the left hand side equals the right hand side. So that's good. The last thing we need to check is that the initial condition is satisfied. So y of zero equals two plus E raise the negative zero cubed. Eat the zero is 1. So we get that Y0 equals three and everything checks out. So that means that it is an exact solution. Okay now for the next part that's just to find the errors that we got for different parts. So we know that why of one equals two plus E. To the negative one cubed. Which industrial is approximately 2.3 679. Right? So the errors are just the difference between the actual answer and the answer regard by using oilers method. So when each equals one we have right? Why of one minus three? Which is about mhm When each equals .1 we have why of one minus two .3928 which equals- .0-4 nine. When e.g. zero one Why of one 2.3 701 equals- 0- two. And then lastly when h equals zero one minus 23681 equals negative 0.2. So those are the errors that are associated with oilers method. And we can see that when H is divided by 10 or error is approximately also divided by 10. Like When you go from H equals .012 h equals .001. Our error goes from zero negative .0022- .0002. So I would say that the air is approximately divided by 10. When H is divided by 10, and that's it.