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course Mth 279
April 12 around 11am.
q_a_08________________________________________
110216
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Question: `q001. Consider the equation
y ' = 3 y^2 * sqrt(t)
This is a separable equation and can be solved. However a very similar equation could be written down which is not solvable. So let's pretend for the moment that this one isn't either.
Suppose we want to calculate an approximate solution curve for which y(1) = .5. Our solution curve will therefore pass through the point (1, .5).
For reference, plot a direction field for this equation on the interval 1 <= t <= 1.2, .5 <= y <= .7.
Now construct an approximate solution by answering the following questions:
What is y ' at the point (1, .5)?
If we follow this slope as we move .1 unit to the right, in the y vs. t plane, ending up at a new point, what will be our rise?
What will then be the coordinates of our new point?
What is y ' at this new point?
If we follow this slope as we move .1 unit to the right, in the y vs. t plane, ending up at a new point, what will be our rise?
What will then be the coordinates of our new point?
We have found two new points by this method. It should be clear that we could continue the process to find as many more new points as we might wish.
Plot these points on your sketch of the direction field. Do they make sense in the context of that field?
The new points lie close to, but not exactly on, the solution curve through (1, .5).
Why do we expect the points to lie close to the solution curve?
Why don't we expect them to lie exactly on the solution curve?
Do our points lie above or below the solution curve? Explain how you know.
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Your solution:
Y’ at (1, 0.5) is 0.75
Rise is 0.075
(1.1, 0.575)
It’s going to be low because of the behavior of 3y^2*sqrt(t).
You just start from (1.1, 0.575) and repeat this process.
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The solution to the equation is straightforward. We get
dy / y^2 = 3 sqrt(t)
which we integrate to get
-1/y = 3 * (2/3 t^(3/2) ) + c
so that
y = - 1 / (2 t^(3/2) + c)
y = .5 when t = 1 yields c = -4, so our solution is
y = -1 / (2 t^(3/2) - 4).
Evaluated at t = 1.1 and 1.2 this yields y values 0.591 and 0.729.
The error in our approximation grows progressively from a discrepancy of .016 at t = 1.1 to .05 at t = 1.2.
The graph below depicts our approximation and the solution curve y = -1 / (2 t^(3/2) - 4).
http://vhcc2.vhcc.edu/dsmith/geninfo/qa_query_etc/differential_equations/q_a_083.gif
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Question: `q002. Solve the equation
y ' = 3 y^2 * sqrt(t)
for the initial condition y(1) = .5.
Evaluate your solution for t = 1.1 and t = 1.2, and plot the resulting two points on your graph of the direction field.
How close was the first new point obtained in the preceding problem to the actual solution?
How close was the second new point obtained in the preceding problem to the actual solution?
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Your solution:
(t, y)
(1.1, 0.5) if you plug this into y’ equation, you get:
= 0.787 SLOPE
(1.2, 0.5) if you plug this into y’ equation, you get:
= 0.822 SLOPE
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Question: `q003. We could get the t = 1.1 approximation by using the same process, with two intervals, each of .05.
What is your approximation to the y value for t = 1.05, based on the slope at (1, .5)?
What is your approximation to the y value for t = 1.10, based on the slope at the point you just found?
How far from the known solution is your new t = 1.10 approximation?
Originally we got to t = 1.1 by a single step. We have now used two steps to get a new approximation.
By what factor did our approximation error change?
Did doubling the number of steps reduce the error in the approximation by a factor of two, less than two or more than two?
Why did the approximation error change as it did?
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Your solution:
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The t = 1.05 approximation is y = .5 + .75 * .05 = .5375.
This yields y ' = 3 * .5375^2 * sqrt(1.05) = .88, approximately, so that the new rise is .88 * .05 = .44 and the t = 1.10 approximation is .5815.
This approximation differs from the accurate t = 1.1 solution .591 by about .0095.
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&#Good responses. See my notes and let me know if you have questions. &#