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course Mth 279
April 28 around 6:30pm.
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:
The actual (unknown) solution starts at (t0,y0) = (0,0)
A starting slope of f(t0,y0) = f(0,0) = 0
Y1 = y0 + f(t0,y0)(t1-t0) = 0 + 0(0) = 0
Step size = h= 0.01
If h = 0.01 and k = 0, 1, 2, 3, then you would just plug each in this expression:
y_(k + 1) = y_k + h f (t_k, y_k)
and solve for y_k (Euler’s method). Right?!
@& Right, but you are also asked to first actually solve the equation. It's a simple linear nonhomogeneous equation and easy enough to solve.
You then need to actually execute the Euler approximation and compare it with the original solution.
The equation is easily solved:
Rearrange to the form y ' + y = t, which is linear nonhomogeneous.
An integrating factor is e^t, yielding
e^t y ' + e^t y = t e^t or
(y e^t) ' = t e^t
with solution
y e^t = integral(t e^t dt) = t e^t - e^t + c.
With y(0) = 0 we get c = 1 to our solution is
y(t) = (t e^t - e^t + 1) / e^t = t - 1 + e^-t.
Our approximation will give us approximate values of y(.01), y(.02), y(.03) and y(.04). The actual values are
.0000498, 0.0001987, 0.0004455, 0.0007894.
Our function is
f(y, t) = y ' = -y + t.
Our initial values are y_0 = 0 and t_0 = 0, and h = .01.
For k = 0 we get
y_1 = y_0 + .01 f (t_0, y_0) = 0 + .01 * 0 = 0.
For k = .1 we get
y_2 = y_1 + .01 f (t_1, y_1) = 0 + .01 f(.01, 0) = 0 + .01 * .01 = .00001.
y_3 = y_2 + .01 f (t_2, y_2) = .0001 + .01 * (.02 - .00001) = .0001 + .01 * .01999 = .0002999.
y_4 = y_3 + .01 f (t_3, y_3) = .000299 + .01 * (.03 - .00299) = .00299 + .02701 = .0058.
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Given Solution:
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Self-critique (if necessary):
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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:
y ' = alpha t + beta and y(t_0) = y_0
y values: -1, -.9, -.81 and -.73
t values: 0, .1, .2, .3
Solution starts at: (t0,y0) and the slope is: f(t0,y0)
Y1 = y0 + f(t0,y0)(t1-t0)
Y2 = y1 + f(t1,y1)(t2-t1)
Etc.
Plug in values and solve, using Euler’s method.
So, y1 = y0 + f(t0,y0)*(0-t0)
@& You don't know alpha and beta. You do know the results of Euler's method.
So you have to turn Euler's method around and find those parameters.
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.
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Given Solution:
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Self-critique (if necessary):
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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:
I don’t understand this. Wouldn’t you have to have “step size (h)” or “k” values to know?!
@& Knowing whether the curve is increasing or decreasing, and knowing the concavity, is sufficient to determine whether we're going to get an overestimate or an underestimate.
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.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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