qa 6

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course Mth 279

2-19 1

Differential Equations Notes 110209`q001. Solve the equation

y ' = sin(t) / y, y(0) = 1.

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y ' = sin(t) / y

y(dy/dt) - sin(t) = 0

y^2/2 - (-cos(t)) = C

= y^2/2 + cos(t) = C, for y(0) = 1

1^2/2 + cos(0) = .5 + 1 = 1.5 = c

c = 1.5

y^2/2 + cos(t) = 1.5

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Over what region(s) of the plane does the existence theorem apply?

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for the subinterval (c, d) containing y_0, t_0. y and t hold true for all real numbers so I believe we can grantee uniqueness for the interval (-infin, +infin) which is subinterval of (a, b) where a

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What is the domain of your solution?

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-infin U +infin

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`q002. In general air resistance has components which depend on v and to v^2, where v is the velocity of an object. If v ' = 5 t + 4 t v + t v^2, what is the solution of the equation for which v(0) = 10?

v ' = -5 t + 4 t v + t v^2 v ' = t(5 + 4v + v^2)

Seperated out

v’/(5+4v+v^2) - t = 0

Int(v’/(5+4v+v^2)), letting u= arctan(v + 2), du = 1/(1 + (v+2)^2) dv

dv = (5+4v+v^2) du, plugging back into our equation

Int((1/(5+4v+v^2))* 5+4v+v^2 du) = Int(1 du) = u = arctan(v + 2)

??????I’m not this is the correct approach but I’ve spent some time on this one and this is the best I can come with???????

v’/(5+4v+v^2) - t = 0, after integrals are taken

arctan(v + 2) - t^2/2 = C

arctan(v+2) = t^2/2 + c

v+2 = tan( t^2/2 + c), finally

v = -2 + tan( t^2/2 + c),solution is

10 = -2 + tan(c)

c = arctan(8) = 1.466

???I don’t believe this is correct, but I can’t see where my error has been?????

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Over what t interval can we expect this solution to exist?

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tan fn is undefined for all values but pi/2 and -pi/2 so any combination that makes t + 1.466 = +-pi/2 will be the only pt to which we expect a solution not to exist

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`q003. Show that the equation

2 y tan(t) dy + (y^2 tan(t) sec (t) + 1 / (2 sqrt(t) ) dt = 0

is of the form M dy + N dt = 0, with M_t = N_y.

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M(t, y) = 2 y tan(t)

N(t, y) = (y^2 tan(t) sec (t) + 1 / (2 sqrt(t))

dM/dt = -2y/cos^2(t)

dN/dy= (2ysin(t)/cos^2(t))/(2`sqrt(t)^2) = 2`sqrt(t)^2*( (2ysin(t)/cos^2(t)))

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Integrate to find the function F(t, y) for which F_t = N, and F_y = M.

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`q004. Consider the equation y ' + p(t) y = q(t) y^3.

Let v(t) = y(t)^m, where the value of m has yet to be determined.

What is dv/dt?

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dv/dt = m*y^(m-1)

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What therefore is dy/dt in terms of v and dv/dt?

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dy/dt = (1/m) y^(m-1) dv/dt

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What is y in terms of v?

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y = v^(1/m)

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Rewrite the original equation in terms of p(t), q(t), v and dv/dt.

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(1/m)v^((1-m)/m) dv/dt + p(t) v^(1/m) = q(t)v^(n/m)

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Choose m so that the v in the right-hand side has power 0.

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m = 1-n

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Now the equation is linear nonhomogeneous order 1.

`q What integrating factor allows us to solve the equation?

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e^(Int( m p(t) dt))

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NOTE: This is an example of a Bernoulli Equation, which is of the form

y ' + p(t) y = q(t) y^n.

This equation differs from a first-order linear homogeneous equation by the factor y^n on the right-hand side.

The general method is to change the variable to v = y^m, then choose the value of m that makes the equation linear.

`q005. A population function P(t) satisfies dP/dt = k P, as long as space and resources are unlimited. That is, the rate of growth is proportional to the population.

If, however, the population grows too much, it approaches the carrying capacity L of its environment, and its rate of growth becomes also proportional to (L - P). Thus the rate of population growth is jointly proportional to P and (L - P).

This gives us the equation

dP/dt = k P ( L - P ).

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This equation is separable. Find its general solution. You will need to use partial fractions.

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P = (LCe^(kLt)/1 + Ce^(kLt))

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If L = 1000, P(0) = 100 and P(1) = 200, what is the value of k?

Using P(0) = 100 we find C, then using P(1) =200 we find k

k = approx. .00008

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Self-critique (if necessary):

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Self-critique rating:

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Question: `q002. Do the points seem to be randomly scattered around the straight line or does there seem to be some nonlinearity in your results?

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Your solution: Most of the points were in a straight line.

confidence rating #$&*:

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Given Solution:

The slope of your line should probably be somewhere between 900 cm/s^2 and 950 cm/s^2. The points should be pretty much randomly scattered about the best possible straight line. Careful experiments of this nature have shown that the acceleration of a system of this nature is to a very high degree of precision directly proportional to the proportion of the weight which is suspended.

STUDENT QUESTION

Is this because the best fit line does not completely go through all the data points, even though it looks linear, or am I plotting wrong??????????????????????????

INSTRUCTOR RESPONSE

The slope you gave in the first problem is the average slope between two actual data points. The line through these two data points is unlikely to be the best fit to the data. You should have sketched the line you think fits the data best, coming as close as possible on the average to the data points (and likely not actually going through any of the data points), then used two points on this line to calculate the slope.

However, whatever straight line you used, it won't go through all of the data points, because they don't all lie on a straight line.

So there is some variation in the straight lines people will sketch in their attempt to approximate the best-fit line, and there is variation in the slopes of these estimated lines.

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Self-critique (if necessary): 2

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Self-critique rating: 2

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Question: `q003. If the acceleration of the system is indeed proportional to the net force on the system, then your straight line should come close to the origin of your coordinate system. Is this the case?

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Your solution: No this is not the case. I was very close to the approximation 900 through 925 cm/s^2.

confidence rating #$&*:

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Given Solution:

If the acceleration of the system is proportional to the net force, then the y coordinate of the straight line representing the system will be a constant multiple of the x coordinate--that is, you can always find the y coordinate by multiplying the x coordinate by a certain number, and this 'certain number' is the same for all x coordinates. The since the x coordinate is zero, the y coordinate will be 0 times this number, or 0.

Your graph might not actually pass through the origin, because data inevitably contains experimental errors. However, if experimental errors are not too great the line should pass very close to the origin.

In the case of this experiment the y-intercept was 12.8. On the scale of the data used here this is reasonably small, and given the random fluctuations of the data points above and below the straight-line fit the amount of deviation is consistent with a situation in which precise measurements would reveal a straight line through the origin.

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Self-critique (if necessary): (2) My points did not pass through the origin due to experimental errors.

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Self-critique rating: 2

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Question: `q003. What is it that causes the system to accelerate more when a greater proportion of the mass is suspended?

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Your solution: Gravitational force exerting on the object.

confidence rating #$&*:

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Given Solution:

The gravitational forces exerted on the system are exerted two objects:

• gravitational force is exerted on the suspended mass, i.e., the part of the original mass that has been removed from the cart and is now suspended

• gravitational force is exerted on the cart (including the masses that remain in it).

The frictional force and the force exerted by the ramp together counter the force of gravity on the cart and the masses remaining in it. The gravitational force on the cart and the masses in it therefore does not affect the acceleration of the system.

However there is no force to counter the pull of gravity on the suspended masses. The net force on the system is therefore just the gravitational force acting which acts on the suspended mass.

The force exerted by gravity on the suspended masses is proportional to the number of suspended masses--e.g, if there are twice as many masses there is twice the force. Thus it is the greater gravitational force on the suspended masses that causes the greater acceleration.

STUDENT QUESTION

I am pretty sure I understand this question. Basically the only thing acting in this equation

to cause a different net force or any type of acceleration is gravity?

INSTRUCTOR RESPONSE

Gravity is the source of the unbalanced force, so in a sense it is the cause. However the net force is different when different proportions of the mass are suspended, and to say that gravity is the cause does not explain the differing accelerations.

The cause of the different accelerations is the configuration of the system.

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Self-critique (if necessary): 2, I guess depending on the type of system configuration will make a great difference in how the system accelerates.

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Self-critique rating: 2

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Question: `q004. This results of this sort of experiment, done with good precision, support the contention that for a given mass the acceleration of a system is indeed proportional to the net force exerted on the system. Other experiments can be done using rubber bands, springs, fans and other nongravitational sources of force, to further confirm this result.

In another sort of experiment, we can see how much force is required on different masses to obtain a certain fixed acceleration. In such experiments we find for example that if the mass is doubled, it requires twice the force to achieve the same acceleration, and that in general the force required to achieve a given acceleration is proportional to the amount of mass being accelerated.

In a certain experiment using the same cart and masses as before, plus several additional identical carts, a single cart is accelerated by a single suspended mass and found to accelerate at 18 cm/s^2. Then a second cart is placed on top of the first and the two carts are accelerated by two suspended masses, achieving an acceleration of 20 cm / s^2. Then a third cart is placed on top of the first to and the three carts are accelerated by three suspended masses, achieving and acceleration of 19 cm/s^2. A fourth cart and a fourth suspended mass are added and an acceleration of 18 cm/s^2 is obtained. Adding a fifth cart in the fifth suspended mass an acceleration of 19 cm/s^2 is obtained. All these accelerations are rounded to the nearest cm/s^2, and all measurements are subject to small but significant errors in measurement.

How well do these results indicate that to achieve a given acceleration the amount of force necessary is in fact proportional to the amount of mass being accelerated?

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Your solution:

confidence rating #$&*:

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Given Solution:

The accelerations obtained are all about the same, with only about a 10% variation between the lowest and the highest. Given some errors in the observation process, it is certainly plausible that these variations are the result of such observation errors; however we would have to have more information about the nature of the observation process and the degree of error to be expected before drawing firm conclusions.

If we do accept the conclusion that, within experimental error, these accelerations are the same then the fact that the second through the fifth systems had 2, 3, 4, and 5 times the mass of the first with 2, 3, 4, and 5 times the suspended mass and therefore with 2, 3, 4, and 5 times the net force does indeed indicate that the force needed to achieve this given acceleration is proportional to the mass of the system.

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Self-critique (if necessary): 0, I will have to do more research in reference to this question. I don’t quite understand.

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Self-critique rating: 0

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Question: `q005. Now we note again that the force of gravity acts on the entire mass of the system when an entire system is simply released into free fall, and that this force results in an acceleration of 9.8 m/s^2. If we want our force unit to have the property that 1 force unit acting on 1 mass unit results in an acceleration of 1 m/s^2, then how many force units does gravity exert on one mass unit?

STUDENT QUESTION

so, normally 1 force unit = 9.8m/s^@ acceleration from gravity?

The force unit has nothing to do with the acceleration of gravity. Gravity does exert forces. First we define our unit of force, then we use the observed acceleration of gravity to figure out how much force gravity exerts.

1 force unit accelerates 1 mass unit at 1 m/s^2.

so force must be equal to 9.8 force units to be 1m/s^2

9.8 force units would accelerate 1 mass unit at 9.8 m/s^2

INSTRUCTOR COMMENT

The action of the gravitational force of Earth on object near its surface acts as follows:

• The gravitational force on any freely falling mass accelerates it at 9.8 m/s^2.

• In particular 1 mass unit would, like anything else, accelerate at 9.8 m/s^2 if in free fall.

Thus the gravitational force accelerates 1 mass unit at 9.8 m/s^2.

Our force unit is defined as follows:

• 1 force unit acting on 1 mass unit results in an acceleration of 1 m/s^2.

So, for example,

• 2 force units acting on 1 mass unit would result in an acceleration of 2 m/s^2.

• 3 force units acting on 1 mass unit would result in an acceleration of 3 m/s^2.

• etc.

We can say the same thing in slightly different words:

• To accelerate 1 mass unit at 2 m/s^2 would require 2 force units.

• To accelerate 1 mass unit at 3 m/s^2 would require 3 force units.

• etc.

Following this pattern, we come to the conclusion that

• To accelerate 1 mass unit at 9.8 m/s^2 would require 9.8 force units.

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Your solution:

Gravity gives one mass unit an acceleration of 9.8 m/s^2. So it will take 9.8 force units to exert or accelerate on one mass unit.

confidence rating #$&*:

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Given Solution:

Since gravity gives 1 mass unit an acceleration of 9.8 m/s^2, which is 9.8 times the 1 m/s^2 acceleration that would be experienced from 1 force unit, gravity must exerted force equal to 9.8 force units on one mass unit.

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Self-critique (if necessary): OK, I understood the instructor’s comments very well.

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Self-critique rating: 3

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Question: `q006. If we call the force unit that accelerates 1 mass unit at 1 m/s^2 the Newton, then how many Newtons of force does gravity exert on one mass unit?

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Your solution:

Gravity gives one mass unit an acceleration of 9.8 m/s^2. The same concept will be for Newton force on a mass unit. So it will take 9.8 Newtons to exert a force on one mass unit.

confidence rating #$&*:

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Given Solution:

Since gravity accelerates 1 mass unit at 9.8 m/s^2, which is 9.8 times the acceleration produced by a 1 Newton force, gravity must exert a force of 9.8 Newtons on a mass unit.

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Self-critique (if necessary): OK

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Self-critique rating: 3

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Question: `q007. The mass unit used here is the kilogram. How many Newtons of force does gravity exert on a 1 kg mass?

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Your solution:

If the mass unit is kilogram, the Newtons force exerted on any mass unit should not make a difference as long as it’s a mass unit. So it will take 9.8 Newtons to exert a force on a mass of 1 kg.

confidence rating #$&*:

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Given Solution:

Gravity exerts a force of 9.8 Newtons on a mass unit and the kg is the mass unit, so gravity must exert a force of 9.8 Newtons on a mass of 1 kg.

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Self-critique (if necessary): OK

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Self-critique rating:

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Question: `q008. How much force would gravity exert on a mass of 8 kg?

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Your solution:

If it takes 9.8 Newtons to exert a force on a mass of 1 kg, it will take (9.8 Newtons x 8 = 78.4 Newtons) to exert a force on a mass of 8 kg.

confidence rating #$&*:

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Given Solution:

Gravity exerts 8 times the force on 8 kg as on 1 kg. The force exerted by gravity on a 1 kg mass is 9.8 Newtons. So gravity exerts a force of 8 * 9.8 Newtons on a mass of 8 kg.

STUDENT COMMENT

I think I did it right but am I not supposed to multiply out 8 * 9.8 Newtons to get 78.6 Newtons? I understand the units wouldn’t cancel, but am I supposed to leave it without “completing” it?

INSTRUCTOR RESPONSE

It's fine to complete it, and that's what you are generally expected to do.

The given solution didn't do that because I wanted to emphasize the logic of the solution without the distraction of the (obvious) arithmetic.

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Self-critique (if necessary): OK

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Question: `q009. How much force would be required to accelerate a mass of 5 kg at 4 m/s^2?

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Your solution: The force required to accelerate a mass of 5 kg at 4 m/s^2 is 5 kg x 4 m/s^2, which is (20 kg m/s^2) or 20 Newtons.

confidence rating #$&*:

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Given Solution:

Compared to the 1 Newton force which accelerates 1 kg at 1 m/s^2, 2e have here 5 times the mass and 4 times the acceleration so we have 5 * 4 = 20 times the force, or 20 Newtons. We can formalize this by saying that in order to give a mass m an acceleration a we must exert a force F = m * a, with the understanding that when m is in kg and a in m/s^2, F must be in Newtons. In this case the calculation would be F = m * a = 5 kg * 4 m/s^2 = 20 kg m/s^2 = 20 Newtons. The unit calculation shows us that the unit kg * m/s^2 is identified with the force unit Newtons.

STUDENT QUESTION

9.8 = 5kg * 4m/s^2=9.8newtons = 20 kg

ok, so force is equal to mass time acceleration,

But I am a little confused as to the 9.8 newtons in question 8...it seems you would incorporate this into the

calculation somehow?

INSTRUCTOR RESPONSE

9.8 Newtons came out of a previous situation in which the acceleration was 9.8 m/s^2. That isn't the acceleration here, so 9.8 Newtons isn't relevant to this question.

9.8 Newtons is the force exerted by gravity on 1 kg.

• This is because 1 Newton of force accelerates 1 kg at 1 m/s^2, and gravity accelerates a mass at 9.8 m/s^2.

• 1 Newton accelerates 1 kg at 1 m/s^2, so 9.8 N are required to accelerate 1 kg at 9.8 m/s^2.

However the number 9.8 doesn't enter into the current situation at all.

We are asked in this question how much force is required to accelerate 5 kg at 4 m/s^2. We could use the following chain of reasoning:

• 1 Newton accelerates 1 kg at 1 m/s^2, so to accelerate 1 kg at 4 m/s^2 requires 4 Newtons.

• 4 Newtons accelerate 1 kg at 4 m/s^2, so to accelerate 5 ig at 4 m/s^2 takes 5 times as much, or 20 Newtons.

Our calculation

5 kg * 4 m/s^2 = 20 Newtons

summarizes this reasoning process.

The result of our reasoning can be generalized:

To get the force required to give a given mass a given acceleration, we multiply the mass by the acceleration.

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Self-critique (if necessary): OK

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Self-critique rating: 3

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Question: `q010. How much force would be required to accelerate the 1200 kg automobile at a rate of 2 m/s^2?

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Your solution:

It will take a force of (1200 kg x 2 m/s^2) to accelerate the automobile.

1200 kg x 2 m/s^2 = 2400 kg m/s^2 or 2400 Newtons.

confidence rating #$&*:

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Given Solution:

This force would be F = m * a = 1200 kg * 2 m/s^2 = 2400 kg * m/s^2 = 2400 Newtons.

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Self-critique (if necessary): OK

Self-critique rating: 3

Questions related to Class Notes

1. An ideal cart of mass 8 kg rolls frictionlessly down an incline. The incline has a length of 50 cm and its 'rise', from low end to high end, is 3 cm. According to the relationship between the accelerating force, the slope and the weight of an object on an incline, as described in the Class Notes:

• What are the magnitude and direction of the force that accelerates the cart down the incline?

• What therefore is the acceleration of the cart?

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

qa 6

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course Mth 279

2-19 3

q_a_06________________________________________

Differential Equations Notes 110209

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Question: `q001. Solve the equation

y ' = sin(t) / y, y(0) = 1.

Over what region(s) of the plane does the existence theorem apply, and what is the domain of your solution?

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Your solution:

y ' = sin(t) / y

y(dy/dt) - sin(t) = 0

y^2/2 - (-cos(t)) = C

= y^2/2 + cos(t) = C, for y(0) = 1

1^2/2 + cos(0) = .5 + 1 = 1.5 = c

c = 1.5

y^2/2 + cos(t) = 1.5

for the subinterval (c, d) containing y_0, t_0. y and t hold true for all real numbers so I believe we can grantee uniqueness for the interval (-infin, +infin) which is subinterval of (a, b) where a

confidence rating #$&*:

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Given Solution:

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Self-critique:

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Self-critique Rating:

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Question: `q003. Show that the equation

2 y tan(t) dy + (y^2 tan(t) sec (t) + 1 / (2 sqrt(t) ) dt = 0

is of the form M dy + N dt = 0, with M_t = N_y.

Integrate to find the function F(t, y) for which F_t = N, and F_y = M.

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Your solution:

M(t, y) = 2 y tan(t)

N(t, y) = (y^2 tan(t) sec (t) + 1 / (2 sqrt(t))

dM/dt = -2y/cos^2(t)

dN/dy= (2ysin(t)/cos^2(t))/(2`sqrt(t)^2) = 2`sqrt(t)^2*( (2ysin(t)/cos^2(t)))

confidence rating #$&*:

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@& This equation is of the given form if M(y) = 2 y tan(t) and N(t) = (y^2 tan(t) sec(t) + 1 / (2 sqrt(t) ) .

M_t = 2 y tan(t) sec(t) and N_y = 2 y tan(t) sec(t)

so the condition holds.

If F_t = N then integral(F_t dt) = integral(N dt) so

F(t) = integral(N dt) = integral( (y^2 tan(t) sec(t) + 1 / (2 sqrt(t) ) dt) = y^2 / cos(t) + sqrt(t) + const.

Since the integral is with respect to t, in which y is treated as a constant, the constant can be any function of y, so

F(t) = y^2 / cos(t) + sqrt(t) + f(y).

Since F_y = M we also get

F(t) = integral(M dy) = integral ( 2 y tan(t) sec(t) dy) = y^2 / cos(t) + g(t), so

F(t) = y^2 / cos(t) + g(t).

We can reconcile the two forms of F(t) by letting f(y) = 0 and g(t) = sqrt(t), so that now

F(t) = y^2 / cos(t) + sqrt(t).

`q Our equation is now of the form dF(t, y) = 0, so that its solution is

F(t, y) = c.

This gives us at least an implicit solution.

The implicit solution F(t, y) = c is

y^2 / cos(t) + sqrt(t) = c.

We note that this solution is defined as long as t >= 0, so that sqrt(t) is defined, and cos(t) is not zero, which eliminates solutions t = pi/2 + n * pi .

We can solve for y to obtain an explicit solution:

y = sqrt(cos(t) sqrt(t) + c cos(t) ),

where c is an arbitrary constant.

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Given Solution:

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Self-critique:

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Self-critique Rating:

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Question: `q004. Consider the equation y ' + p(t) y = q(t) y^3.

Let v(t) = y(t)^m, where the value of m has yet to be determined.

What is dv/dt?

What therefore is dy/dt in terms of v and dv/dt?

What is y in terms of v?

Rewrite the original equation in terms of p(t), q(t), v and dv/dt.

Choose m so that the v in the right-hand side has power 0.

Now the equation is linear nonhomogeneous order 1.

`q What integrating factor allows us to solve the equation?

NOTE: This is an example of a Bernoulli Equation, which is of the form

y ' + p(t) y = q(t) y^n.

This equation differs from a first-order linear homogeneous equation by the factor y^n on the right-hand side.

The general method is to change the variable to v = y^m, then choose the value of m that makes the equation linear.

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Your solution:

dv/dt = m*y^(m-1)cv dv/dt = m*y^(m-1)’

dy/dt = (1/m) y^(m-1) dv/dt

y = v^(1/m)

(1/m)v^((1-m)/m) dv/dt + p(t) v^(1/m) = q(t)v^(n/m)

m = 1-n

e^(Int( m p(t) dt))

confidence rating #$&*:

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@& Using ' for differentiation with respect to t, v = y^m so v ' = (y^m) ' = m y^(m-1) y '.

In terms of dv/dt and dy/dt we have

dv/dt = m y^(m-1) dy/dt.

dv/dt = m y^(m-1) dy/dt so

dy/dt = 1/m * y^(1 - m) * dv/dt.

v = y^m, so y = v^(1/m).

The equation y ' + p(t) y = q(t) y^3 becomes

1/m * y^(1 - m) * dv/dt + p(t) * v^(1/m) = q(t) * v^(3 / m).

This form still includes the y function. Since y = v^(1/m) we get

1/m * v^( (1 - m) / m ) * dv/dt + p(t) * v^(1/m) = q(t) v^(3 / m).

Multiplying through by m v^((m - 1) / m) we get

dv/dt + m p(t) v^(1/m + (m-1)/m) = q(t) v^(3/m + (m - 1) / m), or

dv/dt + m p(t) v = q(t) v^( (m+2) / m).

The power of v on the right-hand side is (m + 2) / m. If m = -2, then the power is 0.

The resulting equation is

dv/dt + 2 p(t) v = q(t).

This equation, which as we have seen is equivalent to the original equation with the substitution v = y^-2, is linear nonhomogeneous order 1.

The coefficient of the linear term (i.e., the v term) is m p(t), so the integrating factor is e^(integral( 2 p(t) ) dt) = e^(2 integral(p(t) dt).

NOTE: This is an example of a Bernoulli Equation, which is of the form

y ' + p(t) y = q(t) y^n.

This equation differs from a first-order linear homogeneous equation by the factor y^n on the right-hand side.

The general method is to change the variable to v = y^m, then choose the value of m that makes the equation linear.

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Given Solution:

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Self-critique:

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Self-critique Rating:

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Question:

`q005. A population function P(t) satisfies dP/dt = k P, as long as space and resources are unlimited. That is, the rate of growth is proportional to the population.

If, however, the population grows too much, it approaches the carrying capacity L of its environment, and its rate of growth becomes also proportional to (L - P). Thus the rate of population growth is jointly proportional to P and (L - P).

This gives us the equation

dP/dt = k P ( L - P ).

This equation is separable. Find its general solution. You will need to use partial fractions.

If L = 1000, P(0) = 100 and P(1) = 200, what is the value of k?

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Your solution:

P = (LCe^(kLt)/1 + Ce^(kLt)), general solution

For P(0) = 100 we have …

P = (LCe^(kLt)/1 + Ce^(kLt)) can be rewritten as P = (L/(Ce^(-kLt) + 1)

P = 1000/(C + 1) = 100

C+1 = 10

C = 9

For P(1) = 200 we have …

P = (LCe^(kLt)/1 + Ce^(kLt)) can be rewritten as P = (L/(Ce^(-kLt) + 1)

P = 1000/(9e^(-1000k) + 1) = 200

9e^(-1000k) + 1 = 5

9e^(-1000k) = 4

e^(-1000k) = 2.25

-1000k = ln(2.25)

k = ln(2.25)/-1000 = approx. -.0008

confidence rating #$&*:

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Given Solution:

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Self-critique:

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Self-critique Rating:

@& We get

dP / (P ( L - P) ) = k dt.

We rewrite 1 / (P ( L - P ) ), using partial fractions:

1 / (P ( L - P ) ) = a / P + b / (L - P) for appropriate values of A and B.

a / P + b / (L - P) = ( a ( L - P ) + b P ) / (P ( L - P )) = (a * L + ( b - a ) * P )

1 / (P ( L - P ) ) = (a * L + ( b - a ) * P ) / (P ( L - P ) ).

The denominators of both sides are equal, so their numerators must be equal. So

a * L + (b - a) * P = 1.

This must be so for all values of P. Thus

a * L = 1 and

(b - a) = 0.

We conclude that a = 1 / L and b = - 1 / L.

Thus

1 / (P ( L - P ) ) = (1 / L) / P + (-1 / L ) / (L - P).

Our equation becomes

(1 / L) / P + (-1 / L ) / (L - P) = k dt

Integrating both sides we get

1 / L * ln | P | - 1 /L * ln | L - P | = k t + c

so that

ln | P | - ln | L - P | = k L t + c

ln (| P | / | L - P |) = k L t + c

| P | / | L - P | = e^(k L t + c).

P and L are both non-negative, so as long as P < L we have

P / (L - P) = A e^(k L t), where A > 0 = e^c can be any positive constant.

Solving for P:

P = (L - P) A e^(k L t)

P = L * A e^(k L t) - P * A e^(k L t)

P ( 1 + A e^(k L t)) = L * A e^(k L t)

P = L * A e^(k L t) / (1 + A e^(k L t) ) = L / (A e^(-k L t) + 1)

Note that as t gets large, e^(- k L t) gets small, so that the denominator approaches 1.

The limiting value of P, as t approaches infinity, is thereofore L / 1 = L.

The t = 0 population is L / (A + 1). If we know the initial population P(0) and the carrying capacity L, then, we can find the value of A.

P(0) = L / (A e^(-k L * 0 ) + 1) = L / (A + 1), so

100 = 1000 / (A + 1).

An easy solution gives us A = 9.

Thus our function is

P(t) = 1000 / (9 e^-(1000 k t) + 1).

We also know that P(1) = 200, so

200 = 1000 / (9 e^-(1000 k) + 1).

Thus

200 * (9 e^-(1000 k) + 1) = 1000

e^(-1000 k) = 800 / 1800 = 4/9

-1000 k = ln(4/9)

k = -ln(4/9) / 1000 = .0009 (instructor's very rough mental estimate, not expected to be very accurate).

Our function is therefore

P(t) = 1000 / (9 e^(- .9 t) + 1)

(again based on mental estimates; the exponent -.9 t is in the right ballpark but not expected to be extremely accurate).

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@& You've got good solutions to most of these problems. Check the inserted solutions.*@