Query 11

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course MTH 279

2/24 4 pm

Query 11 Differential Equations*********************************************

Question:  3.8.4.  Solve the equation y ' = - y + t with y(0) = 0.

Write the expression y_(k + 1) = y_k + h f (t_k, y_k) for h = .01.

Find y_k for k = 0, 1, 2, 3.

Using your original solution for the equation, compare your values of y_k with the values given by the accurate solution.

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Your solution: 

y = t - 1 +e^(-t)

y_(k + 1) = y_k + .01*f (t_k, y_k)

y_1 = 0 + .01[0+0]

=0

y_2 = 0 + .01[0+.01]

=.0001

y_3 = .0001 +.01[-.0001+.02]

=.000299

y_4 = .000596

Actual values:

y_1 = .00005

y_2 = .000199

y_3 = .000446

y_4 = .000789

Euler's method gives an underapproximation to the solution.

 

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary): OK

 

 

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Self-critique rating: OK

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Question:  3.8.6.  Euler's method applied to the equation y ' = alpha t + beta, y(t_0) = y_0, yields y values -1, -.9, -.81 and -.73 at respective t values 0, .1, .2, .3.  Find the values of alpha, beta, t_0 and y_0.

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Your solution: 

alpha = -1

beta = 1

 t_0 = 0 and y_0 = -1

Using the formula for Euler's method,

y_1 = -1 = y_0 + .1(alpha t_0 + beta)

y_2 = -.9 = -1 + .1(alpha(0) + beta)

So we can see that beta = 1

After that, we can solve for alpha. I used the equation y_3 = -.81 = y_2 + .1(alpha*.1 + 1 )

Now, I tried to solve for y_0 and t_0 by using the y_1 equation and another substituted y_2 equation...which gave me the exact same equation... So, I'm pretty sure t_0 is 0 just based on common sense, and of course the y_0 is already known. Here's that work:

-1 = y_0 + .1(-t_0 +1) simplifies to -1 = y_0 -.1t_0 + .1

In the second equation, I subbed y_1 = y_0 + .1(-t_0 +1) into the equation y_2 = -.9 = y_1 + .1(1), which simplified to -1 = y_0 -.1t_0 + .1

So, if t_0 isn't 0, then the only other way to express the solution would be with a free parameter.

And if my solution is correct, my subscripts may be ""off"" by one.

@&

I don't think you actually say what alpha is, but if you got -1 all is well. Everything else looks very good.

Compare with the following:

t starts at 0, at which point y = -1. So t_0 = 0 and y_0 = -1.

Between t = 0 and t = .1, y changes by +.1 so y ' (0) * .1 = .1. Thus y ' (0) = 1. It follows that y ' (0) = alpha * 0 + beta = beta, and that beta is therefore equal to y ' (0), which we have just found to be 1. That is, beta = 1.

Between t = .1 and t = .2, y changes by .09, so y ' (.1) * .1 = .09 and y ' (.1) = .9. Thus y ' (.1) = alpha * .1 + beta = .1 * alpha + 1, and y ' (.1) = .9. It follows that alpha * .1 + 1 = .9 so alpha = -1.

*@

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary): OK

 

 

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Self-critique rating:

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Question:  3.8.8a.  For each of the following situations, will Euler's method overestimate or underestimate the values of the solution to an equation:

· The solution curve is known to be increasing and concave up.

· The solution curve is known to be increasing and concave down.

· The solution curve is known to be decreasing and concave up.

· The solution curve is known to be decreasing and concave down.

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Your solution: 

 underestimation

overestimation

underestimation

overestimation

 I kind of sketched these out and saw how the slope of the actual graph compared to that which would result from using Euler’s method.

@&

Good. Compare with the following explanations:

If the solution curve is increasing and concave up, the derivative is positive and increasing. Applying the derivative at the left-hand endpoint of our approximation interval, we will be using the minimum value of the derivative on that interval, thereby underestimating the rate of change. Our approximation will therefore be an underestimate.

If the solution curve is increasing and concave down, the derivative is positive and decreasing. Applying the derivative at the left-hand endpoint of our approximation interval, we will be using the maximum value of the derivative on that interval, thereby overestimating the rate of change. Our approximation will therefore be an overestimate.

If the solution curve is decreasing and concave up, the derivative is negative and increasing. Applying the derivative at the left-hand endpoint of our approximation interval, we will be using the minimum value of the derivative on that interval, thereby underestimating the rate of change. Our approximation will therefore be an underestimate.

If the solution curve is decreasing and concave down, the derivative is negative and decreasing. Applying the derivative at the left-hand endpoint of our approximation interval, we will be using the maximum value of the derivative on that interval, thereby overestimating the rate of change. Our approximation will therefore be an overestimate.

*@

 

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary): OK

 

 

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Self-critique rating:

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Question:  3.8.14.  y ' = y^2 with y(0) = 1.

Solve the equation.

Perform Euler's Method to approximate the values of the solution on the t interval [0, 1.2] with step size h = .1.

Compare the values you get with the values given by your solution to the equation.

This could be done by hand, but it would take awhile and the probability of an error would be relatively high.  A spreadsheet is recommended.

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Your solution: 

y = 1/(1-t)

 

k t y_k estimate y_k actual

0 0 1 1

1 .1 1.1 1.1 repeating

2 .2 1.221 1.25

3 .3 1.37 1.43

4 .4 1.557797 1.6 repeating

5 .5 1.8 2

6 .6 2.12464 2.5

7 .7 2.576 3.3 repeating

8 .8 3.239652 5

9 .9 4.289186 10

10 1 6.128898 undefined

11 1.1 9.885238 -10

12 1.2 19.65703 -5

 

confidence rating #$&*:

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Given Solution: 

 

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Self-critique (if necessary): OK

 

 

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Self-critique rating: OK

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&#This looks good. See my notes. Let me know if you have any questions. &#