Query 01

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course Mth 279

6/16 11

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) = -2

find antiderivative

-P(t) = 2t

y = c1*e^(2t)

y = c1*e^2t

Sub in initial condition

3 = c1*e^(2(1)

c1 = 3*e^-2

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

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

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Given Solution:

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Self-critique (if necessary):

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

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Your solution:

y' -9y/t^2 = 0

p(t) = -9/t^2

-P(t) = -9/t

y = c1*e^(-9/t)

sub in initial condition

2 = c1*e^(-9/(1))

2 = c1/e^(-9)

c1 = 2*e^9

solution

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

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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

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Your solution:

y' + ( (2t + 1) / (t^2 + t) ) y = 0

p(t) = ( (2t + 1) / (t^2 + t) )

-P(t) = -ln( |t(t+1)| )

y = c1*e^-ln( |t(t+1)| )

1 = c1*e^-ln( |0| )

ln(0) is undefined. Unsure how to proceed.

@&

The given initial condition is impossible, since ln | 0 | is undefined.

*@

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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

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Your solution:

general solution = c1*e^-P(t)

p(t) = sin(3t)

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

y = c1*e^(cos(3t)/3)

initial condition

2 = c1*e^(cos(0)/3)

2 = c1*e^(1/3)

c1 = 2*e^(-1/3)

solution

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

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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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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E

y = c1*e^(-t^3/3)

graph E depicts a cubic function and the solution contains t^3

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y ' - y = 0

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A

y = c*e^t

solution is standard exponential graph

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y' - y / t = 0

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C

y = c*t

solution is linear and graph C has only linear solutions

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y ' - t y = 0

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B

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

graphs B & D are opposite because their solutions are opposite

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y ' + t y = 0

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F

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

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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:

y = c1*e^(-b*t)

2 = c1*e^(-b*1)

c1 = 2/e^(-b) = 2*e^b

sub into second equation

8 = c1*e^(-b*3)

8 = (2*e^b)*e^(-3b)

b = -ln(2)/2

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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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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y'(t)

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

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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 ' ?

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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.

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w = c1*e^t

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

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y = c1*e^t - 2

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

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y' = c1*e^t

c1*e^t - (c1*e^t - 2) = 2

2 = 2 true

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Your solution:

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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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) = y_0

y(0) = 1

y_0 = 1

y = c1*e^(b*t)

1 = c1*e^(b*(0))

1 = c1*e^0

c1 = 1

y = e^(b*t)

point (-1, 0.5)

0.5 = e^(b*(-1))

b = ln(2)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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&#Very good responses. Let me know if you have questions. &#