Query 1

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course MTH 279

1/24 9 pm

Section 2.2*********************************************

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) = -2t

y = Ce^(2t)

y(1) = Ce^2(1) = 3

C = 3/e^2

y = (3/e^2)*e^2t

y = 3 e^(2t - 2)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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

y = Ce^(-9/t)

C = 2e^9

y = (2e^9) * e^(-9/t)

y = 2 * e^(9 - 9/t)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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

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Your solution:

No solution for the initial condition.

P(t) = ln( abs( t^2 + t) )

y = Ce^[- ln (abs (t^2 + t) ]

y = C[(t^2 + t)^-1]

y(0) = C/0 = 1 Which is not true.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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

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Your solution:

P(t) = -(1/3)cos(3t)

y = Ce^(1/3*cos (3t) )

y(0) = 2 = Ce^(1/3*cos (3*0) )

C = 2/[e^(1/3)]

y = 2/[e^(1/3)] * e^(1/3*cos (3t) )

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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.

In general, I just found the solution to the differential equation and thought about the behavior of the graph. If it was an equation I wasn't all that familiar with (like the first equation), I plotted a few points to get an idea of the behavior and how the slope changed.

y ' - t^2 y = 0

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y ' = t^2 y so the magnitude of the slope increases with distance from y axis (i.e., with increasing | t | ) and with distance from x axis (i.e., with increasing | y | ); slope is negative when y is negative

Graph E does not have these characteristics.
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y ' - y = 0

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This equation corresponds to graph A, as you say.

y ' = y so slope is positive above the x axis, negative below; the magnitude of the slope increases with distance from x axis and is the same on any horizontal line.
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y' - y / t = 0

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Good.

y ' = y / t; along any vertical line slope is equal to y, so magnitude of the slope increases as you move away from the x axis along such a line.

The slope increases in magnitude as you approach the y axis.

The slope is positive in the first and third quadrants, negative in the second and fourth.

Only graph C has these characteristics.

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y ' - t y = 0

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Good.

y ' = t y. The slope is positive in the first and third quadrants, negative in the second and fourth.

The magnitude of the slope increases as you move away from the x axis along any vertical line, and as you move away from the y axis along any horizontal line.
Graph B has these characteristics.

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y ' + t y = 0

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Good.

y ' = - t y. The slope is negative in the first and third quadrants, positive in the second and fourth.

The magnitude of the slope increases as you move away from the x axis along any vertical line, and as you move away from the y axis along any horizontal line.
Graph F has these characteristics.

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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:

b = -.5*ln(4)

y = e^(-bt)

2 = e^(-b)

8 = e^(-3b)

Divide those two equations : 4 = e^(-2b)

Solve for b by taking ln of both sides, etc.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): I don't know if I divided properly, but I feel like I understand the concept.

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Self-critique rating: 3

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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 ' ?

w' = y'

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

y(t) = w(t) - 2

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

w' - (w - 2) = 2

w' - w = 0

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Now solve the equation and check your solution:

Solve this new equation in terms of w.

w = Ce^t

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

y = Ce^t - 2

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

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Your solution:

y = Ce^t - 2

I worked the steps in the problem above. I didn't know if that's how you wanted it done or not.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

y_0 = 1

b = 1

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I got the value for y_0 off of the graph. My logic behind b = 1 is that y = e^(bt) and for the graph to look as it does, it must be e^t, which would mean b had to be 1. I can't really justify it more mathematically than that, so I'm not very confident in my answer.

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Self-critique rating: 3

Query 1

@& y ' = t^2 y so the magnitude of the slope increases with distance from y axis (i.e., with increasing | t | ) and with distance from x axis (i.e., with increasing | y | ); slope is negative when y is negative Graph E does not have these characteristics. *@

@& This equation corresponds to graph A, as you say. y ' = y so slope is positive above the x axis, negative below; the magnitude of the slope increases with distance from x axis and is the same on any horizontal line. *@

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

@&
y ' = t^2 y so the magnitude of the slope increases with distance from y axis (i.e., with increasing | t | ) and with distance from x axis (i.e., with increasing | y | ); slope is negative when y is negative

Graph E does not have these characteristics.
*@

@&
This equation corresponds to graph A, as you say.

y ' = y so slope is positive above the x axis, negative below; the magnitude of the slope increases with distance from x axis and is the same on any horizontal line.
*@