Assignment 18

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course MTH 174

10:39 a.m. 7/22/10

Χ͒ȋ{yΟassignment #018

018. `query 18

Cal 2

07-22-2010

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09:53:19

Query problem 11.3.4 (was 10.3.6) Euler y' = x^3-y^3, (0,0), `dx = .2, 5 steps

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RESPONSE -->

I set up the table with P0= (0,0). Then use 'dy = (slope)('dx) with 'dx = .2. and y= yn+'dy where yn is the previous y value. I get the following ordered pairs: P1=(0.2,0), P2=(0.4,0.002), P3=(0.6,0.014), P4=(0.8,0.058), P5=(1,0.16).

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09:53:44

what is your estimate of y(1)?

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RESPONSE -->

y(1)= .16, approximately.

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10:01:04

Describe how the given slope field is consistent with your step-by-step results.

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RESPONSE -->

This answer is consistent in that we can see at x=0.8 a small slope slope beginning to show on the graph which is 0.5, and y=0.06, approximately which is about .1 which is what the graph shows at this point. Then at x=1 we see the slope very noticeably changing to nearly 1. The y value here is 0.2 which is very close to our approximation of 0.16 which could be rounded further to 0.2.

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10:03:36

Is your approximation an overestimate or an underestimate, and what property of the slope field allows you to answer this question?

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RESPONSE -->

It is an underestimate as this method is similar to the LEFT sums a which would also produce an underestimate for this graph because the slope field for our function is increasing. As was shown in figure 11.27 in the text this method will always produce an underestimate of the true vale.

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10:09:12

Query problem 11.3.10 (was 10.3.10) Euler and left Riemann sums, y' = f(x), y(0) = 0

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RESPONSE -->

I touched on this briefly in my previous answer. The Euler method is like the LEFT sum because they both start at P0, or a in LEFT terminology (as opposed to a+1 in the RIGHT sum). Then they step up by multiplying the 'dx by the slope to find 'dy. The LEFT sum does this because the f(x) in that setup would be equivalent to the y' in the Euler method. Figure 11.25 show the stair steps created by the Euler method and it is identical to the LEFT type of steps created by its rectangles.

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10:09:32

explain why Euler's Method gives the same result as the left Riemann sum for the integral

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RESPONSE -->

I just did this above.

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10:18:48

Query problem 11.4.19 (3d edition 11.4.16) (was 10.4.10) dB/dt + 2B = 50, B(1) = 100

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RESPONSE -->

After a lot of work I finally arrived at B = 75e^(2-2t) + 25. The solution is basically found like every other problem in this section that ends up using e to cancel out the ln. This one threw me off because I wasn't solving for C at the right time. I figured it out after a lot of trial and error both on this problem and other problems like it that had the answer in the back. Just before using e to cancel out the ln |B-25|. I solved for C in terms of the equation at that point using B(1)= 1000. This gave me: C= ln 75 +2. Then substituting this value back into my equation and then proceeding using e I had: B-25 = e^(ln 75 +2-2t) Separating the terms on the left using properties of e I then had: B-25= e^(ln 75)(e^2-2t)= 75e^(2-2t). Then adding 25 to both sides I arrived at the particular solution.

Very good.

Rather than thinking in terms of letting e 'cancel' the ln, simply apply the definition of ln as the inverse function of the exponential. The consequence is that e^a = b and ln(b) = | a | are equivalent statements. So to solve e^(2 - 2t) = (B-25) / 75 for t, we rewrite the equation as 2 - 2 t = ln | (B - 25) / 75 |.

The above is sufficient, but in terms of the algebra of the equation, you can also think in terms of applying an inverse function to both sides. Application of a function does in a sense 'cancel' its inverse, but the process should be thought of in terms of the definition of an inverse function. Thinking in this way we apply the natural log function to both sides, obtaining ln( e^(2 - 2t) ) = ln | (B-25) / 75 |, and then applying the inverse function property we get 2 - 2 t = ln | (B - 25) / 75|.

A solution is given below, for comparison:

** We can separate variables.

We get dB/dt = 50 - 2 B, so that dB / (50 - 2B) = dt.

Integrating both sides we obtain

-.5 ln(50 - 2B) = t + c, where c is an arbitrary integration constant.

This rearranges to

ln | 50 - 2B | = -2 (t + c) so that

| 50 - 2B | = e^(-2(t+c)). Since B(1) = 100 we have, for t = 1, 50 - 2B =

-150 so that |50 - 2 B | = 2 B - 50. Thus

2B - 50 = e^(-2(t+c)) and

B = 25 + .5 * e^(-2(t+c)).

If B(1) = 100 we have

100 = 25 + .5 * e^(-2 ( 1 + c) ) so that

e^(-2 ( 1 + c) ) = 150 and

-2(1+c) = ln(150). Solving for c we find that

c = -1/2 * ln(150) - 1.

Thus

B(t) = 25 + .5 e^(-2(t - 1/2 ln(150) - 1))

= 25 + .5 e^(-2t + ln(150) + 2))

= 25 + .5 e^(-2t) * e^(ln(150) * e^2

= 25 + 75 e^2 e^(-2t)

= 25 + 75 e^(-2 (t 1) ).

Note that this checks out:

B(1) = 25 + 75 e^2 e^(-2 * 1)

= 25 + 75 e^0 = 25 + 75 = 100.

Note also that starting with the expression

| 50 - 2B | = e^(-2(t+c)) we can write

| 50 - 2B | = e^(-2c) * e^(-2 t). Since e^(-2c) can be any positive number we replace this factor by C, with the provision that C > 0. This gives us

| 50 - 2B | = C e^(-2 t) so that

50 2 B = +- C e^(-2 t), giving us solutions

B = 25 + C e^(-2t) and

B = 25 C e^(-2t).

The first solution gives us B values in excess of 25; the second gives B values less than 25.

Since B(1) = 100, the first form of the solution applies and we have

100 = 25 + C e^(-2), which is easily solved to give

C = 75 e^2.

The solution corresponding to the given initial condition is therefore

B = 25 + 75 e^-2 e^(-2t), which is simplified to give us

B = 25 + 75 e^(-2(t 1) ). **

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10:19:19

what is your solution to the problem?

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RESPONSE -->

I gave the details above but the solution is: B= 75e^(2-2t) + 25.

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10:21:38

What is the general solution to the differential equation?

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The general solution would be of the form: B= Ce^(A-2t). Where A and C are found using whatever point is given.

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10:23:38

Explain how you separated the variables for the problem.

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I separated by subtracting 2B from both sides and factored out the 2 to just give B, and this point I had: dB/dt= 2(25-B) Then multiplying both sides by dt and then dividing both sides by 25-B I had my separation: 25- B dB= 2 dt.

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10:25:01

What did you get when you integrated the separated equation?

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I had: -ln | 25 - B| = 2t + C.

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10:33:52

Query problem 11.4.40 (3d edition 11.4.39) (was 10.4.30) t dx/dt = (1 + 2 ln t ) tan x, 1st quadrant

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RESPONSE -->

This problem involved a little trickier separation but was fairly straightforward. I divided both sides by t and multiplied by dt, and finally divided by tan x to get: 1/tan x dx= (1 + 2 ln t)/t dt. The left side was integrated by the substitutoin of (cos x)/(sin x) then u sin x and du= cos x dx. This gave me ln |sin x|. The right side I used the substitution u= ln x du= 1/x dx and I ended up with: ln |t| + (ln |t|)^2 + C. I reduced this using some algebra to: x= arcsin (t*e^((ln |t|)^2 + C).

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10:34:05

what is your solution to the problem?

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RESPONSE -->

The solution is given above.

** We separate variables.

t dx/dt = (1 + 2 ln t) tan x is rearranged to give

dx / (tan x) = (1 + 2 ln t) / t * dt = 1/t dt + 2 ln(t) / t * dt or

cos x / sin x * dx = = 1/t dt + 2 ln(t) / t * dt .

Integrating both sides

we let u = sin(x) on the left, obtaining du / u with antiderivative ln u =

ln(sin(x))

we let u = ln(t) and dv = 1/t * dt on the right and use integration by parts

to get antiderivative ln(t)^2 - int(ln(t) / t). Solving int(ln(t) / t) =

ln(t)^2 - int(ln(t) / t) for ln((t) / t we get int(ln(t) / t) = ln(t)^2 / 2.

int(1/t * dt) = ln(t).

Our equation therefore becomes

ln(sin(x)) = ln(t) + 2 * ln(t)^2 / 2 + c so that

sin(x) = e^(ln(t) + ln(t)^2 + c) = e^(ln(t)) * e^(ln(t)^2) * e^c = A * t *

t^(ln(t)) = A * t^1 * t^(ln(t)) = A * t^(1 + ln(t))

so that

x = arcsin(A * t^(1 + ln(t)) ).

This makes sense for t > 0, which gives a real value of ln(t), as long as A *

(t^(1 + ln(t) ) < 1. **

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10:35:53

What is the general solution to the differential equation?

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RESPONSE -->

The general solution is: x= arcsin (t*e^(ln^2 |t|+ C)

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10:36:19

Explain how you separated the variables for the problem.

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RESPONSE -->

I explained this above. Just a two divisions and one multiplication.

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10:37:04

What did you get when you integrated the separated equation?

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RESPONSE -->

Also given above, but was: ln |sin x| = ln |t| + ln^2 |t| + C

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10:38:10

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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This section hasn't dampened my interest in differential equations. I just think the solutions to these equations are interesting.

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

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