Query 1

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course MTH 279

2/5/2012 7:27PMI dont really feel all that great about this submission.. I feel like i'm missing somethings. Can't wait to get your comments so i can clear some of this up.

Section 2.2

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Question: 1. Solve the following equations with the given initial conditions:

1. y ' - 2 y = 0, y(1) - 3

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Your solution:

P(t) = 2 , integral p(t) = 2t , y = Ce^(2t)

Y(1) = C therefor C = 3

So y = 3e^(2t)

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Given Solution:

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Question: 2. t^2 y ' - 9 y = 0, y(1) = 2.

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Your solution:

P(t) = 9 / t^2

Integral p(t) = -9/t^2

Y = Ce^(-9/t^2) , y(1) = C so C = 2

Y= 2e^(-9/t^2)

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Good idea to use p(t) for the coefficient of y.

Usually P(t) (i.e., upper-case P) is taken to be the integral of p(t).

In any case the integral of p(t) = 9 / t^2 is - 9 / t, not -9 / t^2.

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Given Solution:

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Question: 3. (t^2 + t) y' + (2t + 1) y = 0, y(0) = 1.

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Your solution:

Integral P(t) = ln (|t^2 + t|)

Y = C e^(ln (|t^2 + t|))

y = C(|t^2 + t|)

y(0) = C C = 1

y = 1(|t^2 + t|)

=|t^2 + t|

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Given Solution:

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Question: 4. y ' + sin(3 t) y = 0, y(0) = 2.

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Your solution:

P(t) = sin(3t)

-Integral p(t) = cos(3t)/3 + C

Y = Ce^(cos(3t)/3)

Y(0) = C C=2

Y= 2 e^(cos(3t)/3)

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Given Solution:

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Question: 5. Match each equation with one of the direction fields shown below, and explain why you chose as you did.

y ' - t^2 y = 0

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Y’ = t^2 y

E

Chosen by graphing direction field.

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y ' - y = 0

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Y’ = y

A

Graphed direction field.

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y' - y / t = 0

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Y’ = y/t

C

Graphed direction field.

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y ' - t y = 0

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Y’ = ty

B

Graphed direction field.

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y ' + t y = 0

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Y’ = -ty

F

Graphed direction field.

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A

B

C

D

E

F

6. The graph of y ' + b y = 0 passes through the points (1, 2) and (3, 8). What is the value of b?

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Your solution:

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The solution function is y = C e^(-b t).

If you plug in the coordinates of the two points you get two equations, which you can solve for C and b.

The equation corresponding to the second point would be

8 = C e^(-3 b)

What's the equation corresponding to the first point, and how do you then solve the two equations for C and b?

Hint: Start by dividing one equation by the other.

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Given Solution:

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Question:

7. The equation y ' - y = 2 is first-order linear, but is not homogeneous.

If we let w(t) = y(t) + 2, then:

What is w ' ?

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It seems that w(t) = y(t) + 2 is the same as y’ = y + 2

So I would say that w(t) = y’ where ;

w’(t) = y’’ = y’ ( I assume the + 2 will disappear due to differentiation)

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What is y(t) in terms of w(t)?

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I’m not real sure to be honest… But would it be w(t) = t + 2? Where t would = y?

Or would it be;

W’(t) = w(t) + 2?

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If w = y + 2, then y = w - 2.

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What therefore is the equation y ' - y = 2, written in terms of the function w and its derivative w ' ?

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w’(t) - w(t) = 2

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w ' = y ', and w = y + 2.

So the equation would be

w ' = w

or

w ' - w = 0.

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Now solve the equation and check your solution:

Solve this new equation in terms of w.

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The equation is w ' - w = 0.

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I don’t think im understanding…

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Substitute y + 2 for w and get the solution in terms of y.

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Yep.. still not understanding…

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Check to be sure this function is indeed a solution to the equation.

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No comprende’ :(

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Your solution:

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Given Solution:

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Question: 8. The graph below is a solution of the equation y ' - b y = 0 with initial condition y(0) = y_0. What are the values of y_0 and b?

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Your solution:

I’m not grasping this problem either…

This is the direction that I went, but I don’t think that it is correct.

y’ - by = 0 p(t) = b => integral p(t) = b^2/2 so; y = e^(b^2/2)

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You've integrated with respect to b.

b isn't a variable, it's a constant.

t is your variable of integration.

Integrating b with respect to t you get b * t .

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after reading in the book it says that we need to get the equation into the general form of Ce^(p(t)). In this case we have y= Ce^(b^2/2)

so inorder for y(0) to = y_0 we would need C = y_0 and e^(b^2/2) = 1. If we make b = 0 we get

y = y_0(e^(0^2/2) = y_0 (e^0) = y_0 (1) = y_0

I really don’t understand how I’m supposed to find a value for y_0.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Not bad. Check my notes.

I should also insert a set of solutions. If that isn't included, email me. But in any case you should follow up on my notes before looking at that document.

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