query_11

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Good, but I believe the solution to your equations is alpha = 1.

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This of course affects your solution for beta.

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You've set this up well. What you're doing should work. I think there's just a simple arithmetic error in your solution for alpha.

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Your solutions look plausible. They are underestimates, since y^2 is concave down.

In fact the solution to the equation has a vertical asymptote at t = 1.

Since your approximations were underestimates we would expect that the approach of the approximate solution to the asymptote would be a little delayed.

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You can compare your solutions with the following:

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.

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

Your solution:

Confidence rating:

Given 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.

Self-critique (if necessary):

Self-critique rating:

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.

Your solution:

Confidence rating:

Given Solution:

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.

Self-critique (if necessary):

Self-critique rating:

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.

Your solution:

Confidence rating:

Given Solution:

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.

Self-critique (if necessary):

Self-critique rating:

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.

The equation can be written

dy/dt = y^2

and rearranged to the form

dy / y^2 = dt.

Integrating both sides we get

-1/y = t + c.

Solving for y we obtain

y = -1 / (t + c).

The condition y(0) = 1 gives us

-1 / (0 + c) = 1

which solved for c yields

c = -1

so our solution is

y(t) = 1 / (t - 1).

Note that this function is undefined for t = 1, approaching a vertical asymptote as t -> 1.

Now our numerical solution

Your solution:

Confidence rating:

Given Solution:

Self-critique (if necessary):

Self-critique rating:

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